Mahesh

Practicing the Intermediate 2nd Year Maths 2B Textbook Solutions Inter 2nd Year Maths 2B Differential Equations Solutions Exercise 8(d) will help students to clear their doubts quickly.

Intermediate 2nd Year Maths 2B Differential Equations Solutions Exercise 8(d)

I. Solve the following differential equations.

Question 1.
\(\frac{dy}{dx}=-\frac{(12x+5y-9)}{5x+2y-4}\)
Solution:
A non-homogenous equation
\(\frac{dy}{dx}=-\frac{(ax+by-9)}{a’x+b’y-c’}\) where b = -a’
b = -5, a = 5 ⇒ b = -a
(5x + 2y-4)dy = -(12x + 5y-9) dx
(5x + 2y – 4)dy + (12x + 5y – 9) dx = 0
5 (x dy + y dx) + 2y dy – 4 dy + 12x dx – 9 dx = 0
integrating 5xy + y² – 4y + 6x² – 9x = c

Question 2.
\(\frac{dy}{dx}=-\frac{-3x-2y+5}{2x+3y+5}\)
Solution:
b = – 2, a = 2 ⇒ b = -a
(2x + 3y + 5) dy = (- 3x – 2y + 5) dx
2x dy + 3y dy + 5 dy = -3x dx- 2y dx + 5 dx
2(x.dy + y dx) + By dy + 3x dx + 5 dy – 5 dx = 0
Integrating
2xy + \(\frac{3}{2}\)y² + \(\frac{3}{2}\)x² + 5y – 5x = c
4xy + 3y² + 3x² – 10x + 10y = 2c = c’
Solution is
4xy + 3(x² + y²)- 10(x – y) = c

Question 3.
\(\frac{dy}{dx}=\frac{-3x-2y+5}{2x+3y-5}\)
Solution:
\(\frac{dy}{dx}=\frac{-(3x-2y+5)}{2x+3y-5}\)
Here b = – 2, a¹ = 2
∵ b = -a¹
(2x + 3y – 5) dy = (-3x – 2y + 5) dx „
⇒ 2(x dy + y dx) + (3y – 5) dy + (3x – 5) dx – 0
⇒ 2d (xy) + (3y- 5) dy + (3x- 5) dx = 0
Now integrating term by term, we get
⇒ 2 ∫d (xy) + ∫(3y – 5)dy + ∫(3x – 5)dx = 0
⇒ 2xy + 3.\(\frac{y^2}{2}\) – 5y + 3\(\frac{x^2}{2}\) – 5x = \(\frac{c}{2}\)
or) 3x² + 4xy + 3y² – 10x – 10y = c
Which is the required solution.

Question 4.
2(x – 3y + 1) \(\frac{dy}{dx}\) = 4x – 2y + 1
Solution:
(2x – 6y + 2) dy = (4x – 2y + 1) dx
(2x – 6y + 2) dy – (4x – 2y + 1) dx = 0
2 (x dy + y dx) – 6y dy + 2 dy – 4x dx – dx = 0
Integrating
2xy – 3y² – 2x² + 2y – x = c

Question 5.
\(\frac{dy}{dx}=\frac{x-y+2}{x+y-1}\)
Solution:
b = -1, a’ = 1 ⇒ b = -a’
(x + y – 1) dy = (x – y + 2) dx
(x + y – 1) dy = (x – y + 2) dx = 0
(x dy + y dx) + y dy – dy – x dx – 2 dx = 0
integrating
xy + \(\frac{y^2}{2}\) – \(\frac{x^2}{2}\) – y – 2x = c
2xy + y² – x² – 2y – 4x = 2c = c’

Question 6.
\(\frac{dy}{dx}=\frac{2x-y+1}{x+2y-3}\)
Solution:
b = -1, a = 1 ⇒ b = -a’
(x + 2y – 3) dy = (2x – y + 1) dx
(x + 2y – 3) dy – (2x – y + 1) dx = 0
(x dy + y dx) 4- 2y dy – 3 dy – 2x dx – dx = 0
Integrating
xy + y² – x² – 3y – x = c

II. Solve the following differential equations.

Question 1.
(2x + 2y + 3) \(\frac{dy}{dx}\) = x + y + 1
Solution:
\(\frac{dy}{dx}=\frac{x+y+1}{2x+2y+3}\)

Multiplying with 9
6v + log (3v + 4) = 9x + 9c
6(x + y) + log [3(x + y) + 4] = 9x + c
i.e., log (3x + 3y + 4) = 3x – 6y + c

Question 2.
\(\frac{dy}{dx}=\frac{4x+6y+5}{3y+2x+4}\)
Solution:

Multiplying with 64
8v + 9log (8v + 23) = 64x + 64c
8 (2x + 3y) – 64x + 9 log (16x + 24y + 23) = c’
Dividing with 8
2x + 3y – 8x + \(\frac{9}{8}\) log (16x + 24y + 23) = c”
3y – 6x + \(\frac{9}{8}\) log (16x + 24y + 23) = c”
Dividing with 3, solution is 3
y – 2x + \(\frac{3}{8}\) log (16x + 24y + 23) = k

Question 3.
(2x + y + 1) dx + (4x + 2y – 1) dy = 0
Solution:

∫(2 + \(\frac{1}{v-1}\))dv = 3∫dx
2v + log (v – 1) = 3x + c
2v – 3x + log (v – 1) = c
2(2x + y) – 3x + log (2x + y – 1) = c
4x + 2y – 3x + log (2x + y – 1) = c
Solution is x + 2y + log (2x + y – 1) = c

Question 4.
\(\frac{dy}{dx}=\frac{2y+x+1}{2x+4y+3}\)
Solution:

Multiplying with 8
4v + log (4v + 5) = 8x + 8c
4(x + 2y) – 8x + log [4(x + 2y) + 5] = c’
Solution is
4x + 8y – 8x + log (4x + 8y + 5) = c’
8y – 4x + log (4x + 8y + 5) = c’

Question 5.
(x + y – 1) dy = (x + y + 1)dx
Solution:

v – log v = 2x + c
x + y – log (x + y) = 2x – c
(x – y) + log (x + y) = c is the required
solution.

III. Solve the following differential equations.

Question 1.
\(\frac{dy}{dx}=\frac{3y-7x+7}{3x-7y-3}\)
Solution:
Let x = x + h, y = y + k so that \(\frac{dy}{dx}=\frac{dy}{dx}\)


3ln (v – 1) – 3ln (v + 1) – 7ln (v + 1) – 7ln (v – 1)
14ln x – ln c = – 10ln (v + 1) – 4 ln (v – 1)
ln (v + 1)⁵ + ln (v – 1)² + ln x⁷ = ln c
(v +1)⁵. (v – 1)². x⁷ = c
(\(\frac{y}{x}\) + 1)⁵ (\(\frac{y}{x}\) – 1)².x⁷ = c
(y – x)² (y + x)⁵ = c
[y – (x – 1 )]² (y + x – 1 )⁵ = c
Solution is [y-x + 1 ]² (y + x – 1)⁵ = c.

Question 2.
\(\frac{dy}{dx}=\frac{6x+5y-7}{2x+18y-14}\)
Solution:



Multiplying with (3V – 2)(2V + 1)
2 + 18V = A(2V + 1) + B(3V – 2)

2 log (3V- 2)+ log (2V+ 1) = – 3 log X + log c
log (3V – 2)².(2V + 1) + log X³ = log c
log X³(3V – 2)² (2V + 1) = log c
x³(3V – 2)² (2V + 1) = c

Solution is (3y – 2x – 1)² (x + 2y – 2) = 343c = c”.

Question 3.
\(\frac{dy}{dx}=\frac{10x+8y-12}{7x+5y-9}\) = 0
Solution:
\(\frac{dy}{dx}=\frac{10x+8y-12}{7x+5y-9}\) = 0
x = X + h, y = Y + k


5V + 7 = A(V + 2) + B (V + 1)
V = -1 ⇒ 2 = A(-1 + 2) = A ⇒ A = 2
V = -2 ⇒ -3 = B(-2 + 1) = -B, B = 3
∫(\(\frac{2}{(V+1)}+\frac{3}{(V+2)}\))dv = ∫\(\frac{dx}{X}\)
2 log (V + 1) + 3 log (V + 2) = – 5 log X + c
c = 2 log (V + 1) + 3 log (V + 2) + 5 log X
= log (V + 1)². (V + 2)³. X⁵
= log(\(\frac{2}{(V+1)})\))².(\(\frac{3}{(V+2)}\))³. X⁵
= log\(\frac{(Y+X)^2}{X^2}\) \(\frac{(Y+2X)^3}{X^3}\) . X⁵
⇒ (Y + X)² . (Y + 2X)³ = e^c = c’
(Y + 1 – X – 2)² (Y + 1 – 2x – 4)³ = c
Solution is (x + y – 1)² (2x + y – 3)³ = c.

Question 4.
(x – y – 2) dx + (x – 2y – 3) dy = 0
Solution:
Given equation is \(\frac{dy}{dx}=\frac{-x+y+2}{x-2y-3}\)
Let x = X + h, y = Y + k




is the required solution.

Question 5.
(x – y) dy = (x + y + 1) dx
Solution:

Question 6.
(2x + 3y – 8) dx = (x + y – 3) dy
Solution:

Question 7.
\(\frac{dy}{dx}=\frac{x+2y+3}{2x+3y+4}\)
Solution:
Let x = X + h, y = Y + k so that \(\frac{dY}{dX}=\frac{dy}{dx}\)

Choose h and k so that
h + 2k + 3 = 0
2h + 3k + 4 = 0

This is a homogeneous equation

Question 8.
\(\frac{dy}{dx}=\frac{2x+9y-20}{6x+2y-10}\)
Solution:
Given equation is \(\frac{dy}{dx}=\frac{2x+9y-20}{6x+2y-10}\)
Let x = X + h, y = Y + k so that \(\frac{dY}{dX}=\frac{dy}{dx}\)

∴ \(\frac{dY}{dX}=\frac{2X+9Y}{6X+2Y}\)
This is a homogeneous equation

Practicing the Intermediate 2nd Year Maths 2B Textbook Solutions Inter 2nd Year Maths 2B Differential Equations Solutions Exercise 8(c) will help students to clear their doubts quickly.

Intermediate 2nd Year Maths 2B Differential Equations Solutions Exercise 8(c)

I.

Question 1.
Express x dy – ydx = \(\sqrt{x^2+y^2}\) dx in the form F (\(\frac{y}{x}\)) = (\(\frac{dy}{dx}\)).
Solution:
x. dy – y dx = \(\sqrt{x^2+y^2}\) dx
\(\frac{dy}{dx}\) – y = \(\sqrt{x^2+y^2}\)

Question 2.
Express (x – y Tan^-1 \(\frac{y}{x}\))dx + x Tan^-1\(\frac{y}{x}\) dy = 0 in the form F(\(\frac{y}{x}\)) = \(\frac{dy}{dx}\).
Solution:

Question 3.
Express x\(\frac{dy}{dx}\) = y(log y – log x + 1) in the from F (\(\frac{y}{x}\)) = \(\frac{dy}{dx}\)
Solution:
x . \(\frac{dy}{dx}\) = y(log y – log x + 1)
\(\frac{dy}{dx}\) = \(\frac{y}{x}\)(log\(\frac{y}{x}\) + 1)

II. Solve the following differential equations.

Question 1.
\(\frac{dy}{dx}=\frac{x-y}{x+y}\)
Solution:
\(\frac{dy}{dx}=\frac{x-y}{x+y}\)
Put y = vx
\(\frac{dy}{dx}\) = v + x\(\frac{dv}{dx}\)

Question 2.
(x² + y²) dy = 2xy dx
Solution:
(x² + y²) dy = 2xy dx

1 + v² = A(1 – v²) + Bv(1 – v) + Cv(1 + v)
v = 0 ⇒ = A
v = 1 ⇒ 1 + 1 = C(2) ⇒ c = 1
v = -1 ⇒ 1 + 1 = B(-1) (2) ⇒ 2 = -2B
B = -1

Question 3.
\(\frac{dy}{dx}=\frac{-(x^2+3y^2)}{3x^2+y^2}\)
Solution:
\(\frac{dy}{dx}=\frac{-(x^2+3y^2)}{3x^2+y^2}\)
Put y = vx

Multiplying with (v + 1)³
3 + v² = A(v + 1)² + B(v + 1) + C
v = – 1 ⇒ 3 + 1 = C ⇒ C = 4
Equating the co-efficients of v²
A = 1
Equating the co-efficients of v
0 = 2A + B
B = -2A = -2

Question 4.
y²dx + (x² – xy)dy = 0
Solution:

v – log v = log x + log k
v = log v + log x + log k
= log k (vx)
\(\frac{y}{x}\) = log ky
Solution is ky = e^y/x

Question 5.
\(\frac{dy}{dx}=\frac{(x+y)^2}{2x^2}\)
Solution:

Question 6.
(x² – y²)dx – xy dy = 0
Solution:
(x² – y²)dx – xy dy = 0
(x² – y²)dx = xy . dy

Question 7.
(x²y – 2xy²)dx = (x³ – 3x²y)dy
Solution:
(x²y – 2xy²)dx = (x³ – 3x²y)dy

Question 8.
y²dx + (x² – xy + y²) dy = 0
Solution:
y²dx = – (x² – xy + y²)dy

1 – v + v² = A(1 + v²) + (Bv + C)v
v  =  0 ⇒ 1 = A
Equating the co-efficients of v²
1 = A + .B ⇒ B = 0
Equating the co-efficients of v
-1 = C

Question 9.
(y² – 2xy)dx + (2xy – x²)dy = 0
Solution:
(y² – 2xy)dx + (2xy – x²)dy = 0
(2xy – x²)dy = – (y² – 2xy)dx

2v – 1 = A(1 – v) + Bv
v = 0 ⇒ -1 = A ⇒ A = -1
v = 1 ⇒ 1 = B ⇒ B = 1

Question 10.
\(\frac{dy}{dx}+\frac{y}{x}=\frac{y^2}{x^2}\)
Solution:
\(\frac{dy}{dx}+\frac{y}{x}=\frac{y^2}{x^2}\)

Question 11.
x dy – y dx = \(\sqrt{x^2+y^2}\)
Solution:
x dy – y dx = \(\sqrt{x^2+y^2}\)

Question 12.
(2x – y)dy = (2y – x)dx
Solution:


⇒ (y – x)² = c²(y + x)²(y² – x²)
⇒ y – x = c²(y + x)³
⇒ (x + y)³ = c(x – y) where c = \(\frac{-1}{c^2}\)
∴ (x + y)³ = c(x – y)

Question 13.
(x² – y²)\(\frac{dy}{dx}\) = xy
Solution:

Question 14.
2\(\frac{dy}{dx}=\frac{y}{x}+\frac{y^2}{x^2}\)
Solution:
Put y = vx

(y – x)² = y²c²x
Solution is y²x = c(x – y)²

III.

Question 1.
Solve : (1 + e^x/y)dx + e^x/y(1 – \(\frac{x}{y}\))dy = 0.
Solution:
Put x = vy

Question 2.
Solve : x sin \(\frac{y}{x}.\frac{dy}{dx}\) = y sin \(\frac{y}{x}\) – x
Solution:
Dividing with x, we have

Question 3.
Solve : x dy = (y + x cos² \(\frac{y}{x}\))dx
Solution:
x.\(\frac{dy}{dx}\) = y + xcos² \(\frac{y}{x}\)

Question 4.
Solve : (x – y log y + y log x)dx + x(log y – log x)dy = 0
Solution:
Dividing with x. dx we get

Question 5.
Solve : (y dx + x dy) x cos \(\frac{y}{x}\) = (x dy – y dx) y sin \(\frac{y}{x}\).
Solution:
The given equation can be written as

∴ This is a homogeneous equation

Question 6.
Find the equation of a curve whose gradient is \(\frac{dy}{dx}=\frac{y}{x}-\cos^2\frac{y}{x}\), where x > 0, y > 0 and which passes through the point (1, \(\frac{\pi}{4}\)).
Solution:

tan v = – log |x| + c
This curve passes through (1, \(\frac{\pi}{4}\))
tan \(\frac{\pi}{4}\) = c – log 1
c = 1
∴ Equation of the curve is
tan v = 1 – log |x|
tan \(\frac{y}{x}\) = 1 – log |x|

Practicing the Intermediate 2nd Year Maths 2B Textbook Solutions Inter 2nd Year Maths 2B Differential Equations Solutions Exercise 8(b) will help students to clear their doubts quickly.

Intermediate 2nd Year Maths 2B Differential Equations Solutions Exercise 8(b)

I.

Question 1.
Find the general solution of \(\sqrt{1-x^2}\) dy + \(\sqrt{1-y^2}\) dx = 0.
Solution:
Given differential equation is

sin^-1 y = – sin^-1 x + c
Solution is sin^-1 x + sin^-1 y = c, where c is a constant.

Question 2.
Find the general solution of \(\frac{dy}{dx}=\frac{2y}{x}\).
Solution:
\(\frac{dy}{dx}=\frac{2y}{x}\)
∫\(\frac{dy}{dx}\) = 2∫\(\frac{2y}{x}\)
log c + log y = 2 log x
log cy = log x²
Solution is cy = x², where c. is a constant.

II. Solve the following differential equations.

Question 1.
\(\frac{dy}{dx}=\frac{1+y^2}{1+x^2}\)
Solution:
\(\frac{dy}{dx}=\frac{1+y^2}{1+x^2}\)
∫\(\frac{dy}{1+y^2}\) = ∫\(\frac{dx}{1+x^2}\)
tan^-1 y = tan^-1 x + tan^-1c where c is a constant.

Question 2.
\(\frac{dy}{dx}\) = e^y-k
Solution:
\(\frac{dy}{dx}=\frac{e^y}{e^x}\)
\(\frac{dy}{e^y}=\frac{dx}{e^x}\)
∫e^-xdx = ∫e^-ydy
-e^-x = -e^-y + C
e^-y = e^-x + c where c is a constant.

Question 3.
(e^x + 1) y dy + (y + 1) dx = 0
Solution:
(e^x + 1 )y. dy = – (y + 1) dx

y – log (y + 1) = log (e^-x + 1) + log c
⇒ y – log (y + 1) = log c (e^-x + 1)
⇒ y = log (y + 1) + log c (e^-x + 1)
y = log c (y + 1) (e^-x +1)
Solution is
e^y = c(y + 1) (e^-x +1)

Question 4.
\(\frac{dy}{dx}\) = e^x-y + x²e^-y
Solution:
\(\frac{dy}{dx}\) = e^x-y + x² . e^-y
= \(\frac{e^x}{e^y}=\frac{x^2}{e^y}\)
∫e^y . dy = ∫(e^x + x²) dx
Solution is
e^y = e^x + \(\frac{x^3}{3}\) + c

Question 5.
tan y dx + tan x dy = 0
Solution:
tan y dx = – tan x dy

log sin x = – log sin y + log c
log sin x + log sin y = log c
log (sin x . sin y) = log c
⇒ sin x . sin y = c is the solution

Question 6.
\(\sqrt{1+x^2}\)dx + \(\sqrt{1+y^2}\)dy = 0
Solution:
\(\sqrt{1+x^2}\)dx = –\(\sqrt{1+y^2}\)dy
Integrating both sides we get
∫\(\sqrt{1+x^2}\)dx = -∫\(\sqrt{1+y^2}\)dy
Integrating both sides we get

Question 7.
y – x\(\frac{dy}{dx}\) = 5(y² + \(\frac{dy}{dx}\))
Solution:
y – 5y² = (x + 5)\(\frac{dy}{dx}\)
\(\frac{dx}{x+5}=\frac{dy}{y(1-5y}\)
Integrating both sides

Question 8.
\(\frac{dy}{dx}=\frac{xy+y}{xy+x}\)
Solution:

III. Solve the following differential equations.

Question 1.
\(\frac{dy}{dx}=\frac{1+y^2}{(1+x^2)xy}\)
Solution:

log (1 + y²) = log x² – log (1 + x²) + log c
log (1 + x²) + log (1 + y²) = log x² + log c
Solution is (1 + x²) (1 + y²) = cx²

Question 2.
\(\frac{dy}{dx}\) + x² = x² e^3y
Solution:

log(1 – e^-3y) = x³ + c'(c’ = 3c)
Solution is
1 – e^-3y = e^x³ . k(k = e^c’)

Question 3.
(xy² + x)dx+(yx²+y)dy = 0.
Solution:
(xy² + x) dx + (yx² + y) dy = 0
x(y² + 1) dx + y (x² + 1) dy = 0
Dividing with (1 + x²) (1 + y²)
\(\frac{x dx}{1+x^2}+\frac{y dy}{1+y^2}\) = 0
Integrating
∫\(\frac{x dx}{1+x^2}\) + ∫\(\frac{y dy}{1+y^2}\) = 0
\(\frac{1}{2}\) [(log (1 + x²) + log (1 + y²)] = log c
log (1 + x²) (1 + y²) = 2 log c = log c²
Solution is (1 + x²) (1 + y²) = k when k = c².

Question 4.
\(\frac{dy}{dx}\) = 2y tanh x
Solution:
\(\frac{dy}{dx}\) = 2y tanh x
\(\frac{dy}{y}\) = 2 tanh x dx
Integrating both sides we get
∫\(\frac{dy}{y}\) = 2 ∫ tanh x dx
log y = 2 log |cosh x| + log c
lny = 2ln cosh x + In c
y = c cos²h x

Question 5.
sin^-1 \(\frac{dy}{dx}\) = x + y
Solution:
\(\frac{dy}{dx}\) = sin(x + y)
x + y = t
1 + \(\frac{dy}{dx}=\frac{dt}{dx}\)
\(\frac{dt}{dx}\) – 1 = sin t
\(\frac{dt}{dx}\) = 1 + sin t
Integrating both sides we get
∫\(\frac{dt}{1+\sin t}\) = ∫dx
∫\(\frac{1-\sin t}{\cos^2 t}\) dt = x + c
∫sec² t dt – ∫tan t . sec t dt = x + c
tan t – sec t = x + c
⇒ tan (x + y) – sec (x + y) = x + c

Question 6.
\(\frac{dy}{dx}+\frac{y^2+y+1}{x^2+x+1}\) = 0
Solution:
\(\frac{-dy}{y^2+y+1}=\frac{dx}{x^2+x+1}\)
Integrating both sides dy

Question 7.
\(\frac{dy}{dx}\) = tan² (x + y)
Solution:
\(\frac{dy}{dx}\) = tan² (x + y)
put v = x + y

2v + sin 2v = 4x + c’
2(x + y) + sin 2(x + y) = 4x + c’
x – y – \(\frac{1}{2}\)sin [2(x + y)] = c

Practicing the Intermediate 2nd Year Maths 2B Textbook Solutions Inter 2nd Year Maths 2B Differential Equations Solutions Exercise 8(a) will help students to clear their doubts quickly.

Intermediate 2nd Year Maths 2B Differential Equations Solutions Exercise 8(a)

I.

Question 1.
Find the order of the differential equation obtained by eliminating the arbitrary constants b and c from xy = ce^x – be^-x + x².
Solution:
Given equation is xy = ce^x – be^-x + x²
Differentiating w.r.to x, we get
xy₁ + y = ce^x – be^-x + 2x.
Again differentiating w.r.to x, we get
xy₂ + y₁ + y₁ = ce^x – be^-x + 2
xy₂ + 2y₂ = xy – x² + 2
Arbitary constants a and b are eliminated.
∴ The order is 2.

Question 2.
Find the order of the differential equation of the family of all circles with their centres at the origin.
Solution:
Equation of the circle with centre at origin is x² + y² = r²
Order = no .of arbitrary constants = 1

II.

Question 1.
Form the differential equations of the following family of curves where parameters are given in brackets.
i) y = c(x – c)² ; (c)
Solution:
y = c(x – c)² ………….. (1)
Differentiating w.r. to x
y₁ = c. 2(x – c) ………….. (1)
Dividing (2) by (1)
\(\frac{y_2}{y}=\frac{2c(x-c)}{c(x-c)^2}\)
x – c = \(\frac{2y}{y_1}\)
c = x – \(\frac{2y}{y_1}\)
Substituting in (1)
y = x – \(\frac{2y}{y_1}\)(x – \(\frac{2y}{y_1}\))²
= \(\frac{xy_1-2y}{y_1}.\frac{4y^2}{y_1^2}\)
y.y³₁ = 4y²(xy₁ – 2y)
i.e., y³₁ = 4y (xy₁ – 2y)
= 4xyy₁ – 8y²
(\(\frac{dy}{dx}\))³ – 4xy\(\frac{dy}{dx}\) + 8y² = 0

ii) xy = ae^x + be^-x; (a, b)
Solution:
xy = ae^x + b.e^-x
Differentiating w.r.t. x
x . y₁ + y = ae^x – b . e^-x
Differentiating again w.r.t. x
xy₂ + y₁ + y₁ = ae^x + be^-x = xy
\(\frac{d^2y}{dx^2}\) + 2\(\frac{dy}{dx}\) – xy = 0

iii) y = (a + bx)e^kx ; (a, b)
Solution:
y = (a + bx)e^kx
Differentiating w.r.t. x
y₁ = (a + bx) e^kx. k + e^kx . b
= k . y + b.e^kx
y₁ – ky = b.e^kx …………. (1)
Differentiating again w.r.t. x
y₂ – ky₁ = kb e^kx
= k(y₁ – ky) ………… (2)
= ky₁ – k²y
\(\frac{d^2y}{dx^2}\) – 2k\(\frac{dy}{dx}\) + k²y = 0

iv) y = a cos (nx + b); (a, b)
Solution:
y = a cos (nx + b)
y₁ = – a sin (nx + b) n
y₂ = – an. cos (nx + b) n
= – n² . y
\(\frac{d^2y}{dx^2}\)+n².y = 0

Question 2.
Obtain the differential equation which corresponds to each of the following family of curves.
i) The rectangular hyperbolas which have the co-ordinate axes as asymptotes.
Solution:
Equation of the rectangular hyperbolas is xy = c² where c is arbitrary constant
Differentiating w.r.t. x
x\(\frac{dy}{dx}\) + y = 0

ii) The ellipses with centres at the origin and having co-ordinate axes as axes.
Solution:
Equation of ellipse is
\(\frac{x^2}{a^2}+\frac{y^2}{b^2}\) = 1
Differentiating w.r.to ‘x’ we get

Multiply (ii) by x and subtract from (i)

III.

Question 1.
Form the differential equations of the following family of curves where parameters are given in brackets :
i) y = ae^3x + be^3x; (a, b)
Solution:
Differentiating w.r. to x
y₁ = 3ae^3x + 4be^4x
y₁ – 3a. e^3x = 4b.e^4x
= 4(y – a. e^3x)
= 4y – 4a. e^3x
y₁ – 4y = – a.e^3x ………… (1)
Differentiating again w.r.t. x
y₂ – 4y₁ = – 3a. e^3x
= 3 (y₁ – 4y) by (1)
= 3y₁ – 12y
\(\frac{d^2y}{dx^2}\) – 7\(\frac{dy}{dx}\) + 12y = 0

ii) y = ax² + bx; (a, b)
Solution:

Adding all three equations we get
x²\(\frac{d^2y}{dx^2}\) – 2x\(\frac{dy}{dx}\) + 2y = 0

iii) ax² + by² = 1; (a, b)
Solution:
ax² + by² = 1
by² = 1 – ax² ………….. (1)
Differentiating w.r.t. x
2by. y₁ = – 2ax ………….. (2)
Dividing (2) by (1)
\(\frac{by.y_1}{by^2}=\frac{-ax}{1-ax^2}\)

Differentiating w.r.t. x

iv) xy = ax² + \(\frac{b}{x}\); (a, b)
Solution:
xy = ax² + \(\frac{b}{x}\)
x²y = ax³ + b
Differentiating w.r.t. x
x²y₁ + 2xy = 3ax²
Dividing with x
xy₁ + 2y = 3ax ………… (1)
Differentiating w.r.t. x
xy₂ + y₁ + 2y₁ = 3a
xy₂ + 3y₁ = 3a ………… (2)
Dividing (1) by (2)
\(\frac{xy_1+2y}{xy_2+3y_1}=\frac{3ax}{3a}=x\)
Cross multiplying
xy₁ + 2y = x²y₂ + 3xy
x²y₂ + 2xy₁ – 2y = 0
x²\(\frac{d^2y}{dx^2}\) + 2x\(\frac{dy}{dx}\) – 2y = 0

Question 2.
Obtain the differential equation which corresponds to each of the following family of curves.
i) The circles which touch the Y – axis at the origin.
Solution:
Equation of the required circle is
x² + y² + 2gx = 0
x² + y² = – 2gx …………. (1)
Differentiating w.r. t x
2x + 2yy₁ = – 2g ……….. (2)
Substituting in (1)
x² + y² = x(2x + 2yy₁) by (2)
= 2x² + 2xyy₁
yy² – 2xyy₁ – 2x² = 0
y² – x² = 2xy\(\frac{dy}{dx}\)

ii) The parabolas each of which has a latus rectum 4a and whose axes are parallel to X – axis.
Solution:

Equation of the required parabola is
(y – k)² = 4a (x – h) …………. (1)
Differentiating w.r.t. x
2(y – k)y₁ = 4a …………. (2)
Differentiating w.r.t. x
(y – k) y₂ + y²₁ = 0 …………. (3)
From (2), y – k = \(\frac{2a}{y_1}\)
Substituting in (3)
\(\frac{2a}{y_1}\).y₂ = y²₁ = 0
2ay₂ + y³₁ = 0

iii) The parabolas having their foci at the origin and axis along the X – axis.
Solution:
Equation of parabola be y² = 4a(x + a)

Practicing the Intermediate 2nd Year Maths 2B Textbook Solutions Inter 2nd Year Maths 2B Definite Integrals Solutions Exercise 7(d) will help students to clear their doubts quickly.

Intermediate 2nd Year Maths 2B Definite Integrals Solutions Exercise 7(d)

I.

Question 1.
Find the area of the region enclosed by the given curves.
i) y = cos x, y = 1 – \(\frac{2x}{\pi}\)
Solution:
Equations of the given curves are
y = cos x ………….. (1)
y = 1 – \(\frac{2x}{\pi}\) ………….. (2)
Eliminating y from eq’s (1) and (2)
cos x = 1 – \(\frac{2x}{\pi}\)
When x = \(\frac{\pi}{2}\), cos x = cos\(\frac{\pi}{2}\) = 0
1 – \(\frac{2}{\pi}\), x = \(\frac{2}{\pi}\) . \(\frac{\pi}{2}\) = 1 – 1 = 0
When x = 0, cos x = cos 0 = 1
1 – \(\frac{2x}{\pi}\) = 1 – 0 = 1
∴ Point of intersection are A = (\(\frac{\pi}{2}\), 0) B = [π – 1]

Question 2.
y =cos x, y = sin 2x, x = 0, x = \(\frac{\pi}{2}\).
Solution:

Question 3.
y = x³ + 3, y = 0, x = -1, x = 2.
Solution:
Required area PABQ

Question 4.
y = e^x, y = x, x = 0, x = 1.
Solution:

Question 5.
y = sin x, y = cos x; x = 0, x = \(\frac{\pi}{2}\).
Solution:

Between 0 and \(\frac{\pi}{4}\).
cos x > sin x
Between \(\frac{\pi}{4}\) and \(\frac{\pi}{2}\)
cos x < sin x
Required area

Question 6.
x = 4 – y², x = 0.
Solution:
The given parabola x = 4 – y² meets, the x – axis at A(4, 0) and Y – axis at P(0, 2) and Q(0, -2).

The parabola is symmetrical about X – axis

Required area = 2 Area of OAP

Question 7.
Find the area enclosed with in the curve |x| + |y| = 1
Solution:

II.

Question 1.
x = 2 – 5y – 3y², x = 0.
Solution:
Solving the equation of given curves
2 – 5y – 3y² = 0
3y² + 5y – 2 = 0
(y + 2)(3y – 1) = 0 y = -2 or \(\frac{1}{3}\)

Question 2.
x² = 4y, x = 2, y = 0.
Solution:

Question 3.
y² = 3x, x = 3.
Solution:

The parabola is symmetrical about X – axis
Required area = 2\(\int_0^3\)√3. √x dx

Question 4.
y = x², y = 2x.
Solution:
Given equation are y = x² ………….. (1)
y = 2x …………. (2)

Eliminating y, we get x² = 2x
x² – 2x = 0
x(x – 2) = 0
x = 0 or x = 2
y = 0, or y = 4

Point of intersection are O(0, 0), A(2, 4)

Question 5.
y = sin 2x, y = √3 sin x, x = 0, x = \(\frac{\pi}{6}\).
Solution:
Given equation are y = sin 2x ………… (1)
y = √3 sin x …………. (2)
sin 2x = √3 sin x
2 sin x. cos x = √3 sin x
sin x = 0 or 2 cos x = \(\frac{\sqrt{3}}{2}\)
x = 0, cos x = \(\frac{\sqrt{3}}{2}\) ⇒ x = \(\frac{\pi}{6}\)

Question 6.
y = x², y = x³.
Solution:
Given equations are y = x² ………….. (1)
y = x³ ………. (2)

From equation (1) and (2) x² = x³
x³ – x² = 0
x²(x – 1) = 0
x = 0 or 1

Question 7.
y = 4x – x², y = 5 – 2x
Solution:

y = 4x – x² ………… (i)
y = 5 – 2x …………..(ii)
y = -([x – 2]²) = 4
y – 4 = -(x – 2)²
Solving equations (i) and (ii) we get
4x – x² = 5 – 2x
x² – 6x + 5 = 0
(x – 5)(x – 1 ) = 0
x = 1, 5

Question 8.
Find the area in Sq. units bounded by the x – axis, part of the curve y = 1 + \(\frac{8}{x^2}\) and the ordinates x = 2 and x = 4.
Solution:
Given equations y = 1 + \(\frac{8}{x^2}\)

Question 9.
Find the area of the region bounded by the parabolas y² = 4x and x² = 4y.
Solution:
Equations of the given curve are
y² = 4x
x² = 4y
(\(\frac{x^2}{4}\))² = 4x
\(\frac{x^4}{16}\) = 4x
x⁴ = 64x ⇒ x⁴ = 0 or x³ = 64, x = 4

Question 10.
Find the area bounded by the curve y = lnx the X – axis and the straight line x = e.
Solution:
Equation of the curve is y = lnx
x = 1 ⇒ y = 0
The curve y = lnx meets
X – axis at C(1, 0)

Required area = \(\int_1^e\)lnx dx
= (x.lnx)^e₁ – \(\int_1^e\)x.\(\frac{1}{x}\) dx
= (e.ln e – 1.ln 1) – (x)^e₁
= e – (e – 1)
= e – e + 1 = 1 Sq.unit.

III.

Question 1.
y = x² + 1, y = 2x – 2, x = -1, x = 2.
Solution:
Equation of the curve are
y = x² + 1 …………. (1)
y = 2x – 2 ………….. (2)

Area between the given curves

Question 2.
y² = 4x, y² = 4(4 – x).
Solution:
Equations of the curve are y² = 4x ………… (1)
y² = 4(4 – x) …………. (2)
Eliminating y, we get
4x = 4(4 – x)
2x = 4 ⇒ x = 2
Substituting in equation (1), y² = 8
y = ± 2√2

Points of intersection are
A(2, 2√2), B(2, -2√2)
Required area is symmetrical about X – axis
Area OACB

Question 3.
y = 2 – x², y = x².
Solution:

y = 2 – x² …………. (1)
y = x² …………. (2)
x² = -(y – 2)
From equation (2)
2 – x² = x²
2 = 2x² or x² = 1
x = ±1
Area bounded by two curve be

Question 4.
Show that the area enclosed between the curve y² = 12(x + 3) and y² = 20(5 – x) is 64\(\sqrt{\frac{5}{3}}\).
Solution:
Equation of the curve are
y² = 12(x + 3) ……….. (1)
y² = 20(5 – x) ……….. (2)
Eliminating y
12(x + 3) = 20(5 – x)
3x + 9 = 25 – 5x
8x = 16
x = 2
y² = 12(2 + 3) = 60
y = √60 = ±2√15

Points of intersection are B’ (2, 2√15)
B’ (+2, -2√15)
The required area is symmetrical about X – axis
Area ABCB’

Question 5.
Find the area of the region {(x, y)/x² – x – 1 ≤ y ≤ -1}
Solution:

Question 6.
The circle x² + y² = 8 is divided into two parts by the parabola 2y = x². Find the area of both the parts.
Solution:
Equations of the curves are
x² + y² = 8 ………… (1)
2y = x² ………… (2)
Eliminating y between equations (1) and (2)

Let x² = t
4t + t² = 32
t² + 4t – 32 = 0
(t + 8)(t – 4) = 0
t = -8 (not possible) x² = 4 ⇒ x = ±2

As curve is symmetric about Y – axis, total area be

Question 7.
Show that the area of the region bounded \(\frac{x^2}{a^2}+\frac{y^2}{b^2}\) = 1 (ellipse) is π ab. also deduce the area of the cricle x² + y² = a².
Solution:

The ellipse is symmetrical about X and Y axis
Area of the ellipse = 4 Area of CAB
= 4.\(\frac{\pi}{4}\) ab
Equation of elliple = \(\frac{x^2}{a^2}+\frac{y^2}{b^2}\) = 1

Substituting b = a, we get the circle
x² + y² = a²
Area of the circle = πa(a) = πa² sq. units.

Question 8.
Find the area of region enclosed by the curves y = sin πx, y = x² – x, x = 2.
Solution:
Required area

Question 9.
Let AOB be the positive quadrant of the ellipse \(\frac{x^2}{a^2}+\frac{y^2}{b^2}\) = 1 with OA = a, OB = b. Then show that the area bounded the chord AB and the arc AB of the elliple is \(\frac{(\pi-2) a b}{4}\).
Solution:
Let OA = a, OB = b
Equation of AB is \(\frac{x}{a}+\frac{y}{b}\) = 1
\(\frac{y}{b}\) = 1 – \(\frac{x}{a}\)
y = b(1 – \(\frac{x}{a}\))

Question 10.
Prove that curves y² = 4x and x² = 4y divide the area of the square bounded by the lines x = 0, x = 4, y = 4 and x = 0 into three equal parts.
Solution:

The given equations are y² = 4x …………. (1)
x² = 4y …………. (2)
The points of intersection are O(0, 0) A(4, 4)

Practicing the Intermediate 2nd Year Maths 2B Textbook Solutions Inter 2nd Year Maths 2B Definite Integrals Solutions Exercise 7(c) will help students to clear their doubts quickly.

Intermediate 2nd Year Maths 2B Definite Integrals Solutions Exercise 7(c)

I. Evaluate the following definite integrals.

Question 1.
\(\int_{\pi/2}^{\pi/2}\)sin¹⁰ x dx
Solution:

Question 2.
\(\int_0^{\pi/2}\)cos¹¹ x dx
Solution:

Question 3.
\(\int_0^{\pi/2}\)cos⁷ x . sin²x dx.
Solution:

Question 4.
\(\int_0^{\pi/2}\)sin⁴ x . cos⁴ x dx.
Solution:

Question 5.
\(\int_0^{\pi/2}\)sin³ x cos⁶ x dx.
Solution:
\(\int_0^{\pi/2}\)sin³ x cos⁶ x dx.
\(\int_0^{\pi/2}\)(1 – cos² x) cos⁶ x.sin x dx

Question 6.
\(\int_0^{2\pi}\)sin² x cos⁴ x dx.
Solution:

Question 7.
\(\int_{-\pi/2}^{\pi/2}\)sin² θ cos⁷ θ dθ.
Solution:
sin² θ cos⁷ θ is even function
f(θ) = sin² θ . cos⁷ θ dθ
f(-θ) = sin² (-θ) . cos⁷ (-θ)
= f(θ)
= 2\(\int_{-\pi/2}^{\pi/2}\)sin² θ cos⁷ θ dθ

\(\int_0^{\pi/2}\)sin^m x cosⁿx dx
n is odd n = 7

Question 8.
\(\int_{-\pi/2}^{\pi/2}\)sin³ θ .cos³ θ dθ.
Solution:
f(θ) = sin³ θ . cos³ θ dθ
f(-θ) = sin³ (-θ) . cos³ (-θ)
= -sin³ θ cos³ θ = -f(θ)
f(θ) is odd
∴ \(\int_{-\pi/2}^{\pi/2}\)sin³ θ.cos³ θ dθ = 0

Question 9.
\(\int_0^a\)x(a² – x²)^7/2 dx
Solution:
x = a sin θ, a = a sin θ
dx = a cos θ dθ, θ = π/2
= \(\int_0^{\pi/2}\)a sin θ(a² – a²sin²θ)^7/2 a cos θ dθ
= \(\int_0^{\pi/2}\)a⁹ cos⁸ θ sin θ dθ
= a⁹\(\int_0^{\pi/2}\)cos⁸ . θ sin θ dθ

Question 10.
\(\int_0^2\)x^3/2.\(\sqrt{2-x}\)dx
Solution:
x = 2 cos² θ
dx = 4 cos θ sin θ dθ

II. Evaluate the following integrals.

Question 1.
\(\int_0^1\)x⁵(1 – x)^3/2 dx
Solution:
x = sin² θ
dx = 2 sin θ . cos θ . dθ

Question 2.
\(\int_0^4\)(16 – x²)^5/2 dx
Solution:

Question 3.
\(\int_{-3}^3\)(9 – x²)^3/2 x dx
Solution:
Let f(x) = (9 – x²)^3/2_x
f(x) = (9 – (-x²))^3/2(-x)
= (9 – x²)^3/2 . x
= -f(x)
∴ f is odd function
∴ \(\int_{-3}^3\)(9 – x²)^3/2 x dx = 0

Question 4.
\(\int_0^5\)x³(25 + x²)^7/2 dx
Solution:
Let I = \(\int_0^5\)x³(25 + x²)^7/2 dx
Put x = 5 sin θ
dx = 5 cosθ dθ

Question 5.
\(\int_{-\pi}^{\pi}\)sin⁸ x cos⁷ x dx
Solution:
Let f(x) = sin⁸ x. cos⁷ x
f(-x) = sin⁸ (-x) . cos⁷ (-x)
= sin⁸ x. cos⁷ x
∴ f is even function.
∴ \(\int_{-\pi}^{\pi}\)sin⁸ x cos⁷ x dx = 2\(\int_0^{\pi}\)sin⁸ x cos⁷ x = 0

Question 6.
\(\int_3^7 \sqrt{\frac{7-x}{x-3}}\)dx
Solution:
Put x = 3 cos²θ + 7 sin²θ
dx = (7 – 3)sin2θ dθ
dx = 4 sin 2θ dθ
U.L.
x = 3 cos²θ + 7 sin²θ
7 = 3 cos²θ + 7 sin²θ
4 cos²θ = 0
θ = \(\frac{\pi}{2}\)
L.L
x = 3 cos²θ + 7 sin²θ
3 = 3 sin²θ + 7 sin²θ
4 sin²θ = 0
sinθ = 0
θ = 0
7 – x = 7 – (3 cos²θ + 7 sin²θ)
= (7 – 3)cos²θ
= 4 cos²θ
x – 3 = 3 cos²θ + 7 sin²θ – 3
= (7 – 3)sin²θ
= 4 sin²θ

Question 7.
\(\int_2^6\sqrt{(6-x)(x-2)}\)dx
Solution:
Put x = 2 cos²θ + 6 sin²θ
dx = (6 – 2) sin2θ dθ
dx = 4 sin2θ dθ
U.L
x = 2 cos²θ + 6 sin²θ
6 = 2 cos²θ + 6 sin²θ
4 cos²θ = 0
cos θ = 0
θ = \(\frac{\pi}{2}\)

L.L
x = 2 cos²θ + 6 sin²θ
2 = 2 cos²θ + 6 sin²θ
4 sin²θ = 0
θ = 0
6 – x = 6 – (2 cos²θ + 6 sin²θ)
= (6 – 2) cos²θ
= 4 cos²θ
x – 2 = 2 cos²θ + 6 sin²θ – 2
= (6 – 2)sin²θ
= 4 sin²θ

Question 8.
\(\int_0^{\pi}\)tan⁵x cos⁸x dx
Solution:

III. Evaluate the following integrals.

Question 1.
\(\int_0^1\)x^7/2 (1 – x)^5/2 dx
Solution:
Put x = sin²θ
dx = 2 sin θ cos θ dθ
U.L
x = sin²θ
1 = sin²θ
θ = \(\frac{\pi}{2}\)

L.L
x = sin²θ
0 = sin²θ
θ = 0

Question 2.
\(\int_0^{\pi}\)(1 + cos x)³ dx
Solution:

Question 3.
\(\int_4^9\frac{dx}{\sqrt{(9 – x)(x – 4)}}\)
Solution:
Put x = 4 cos²θ + 9 sin²θ
dx = (9 – 4)sin2θ dθ
dx = 5 sin2θ dθ

U.L
x = 4 cos²θ + 9 sin²θ
9 = 4 cos²θ + 9 sin²θ
5 cos²θ = 0
θ = \(\frac{\pi}{2}\)

L.L
x = 4 cos²θ + 9 sin²θ
4 = 4 cos²θ + 9 sin²θ
5 sin²θ = 0
θ = 0

9 – x = 9 – (4 cos²θ + 9 sin²θ)
= (9 – 4) cos²θ
= 5 cos²θ

x – 4 = 4 cos²θ + 9 sin²θ – 4
= (9 – 4) sin²θ
= 5 sin²θ

Question 4.
\(\int_0^5\)x²(\(\sqrt{5-x}^7\) dx
Solution:
Put x = 5 sin²θ
dx = 10 sinθ cosθ dθ

U.L
x = 5 sin²θ
5 = 5 sin²θ
sin²θ = 1
θ = \(\frac{\pi}{2}\)

L.L
x = 5 sin²θ
0 = 5sin²θ
sin²θ = 0
θ = 0

Question 5.
\(\int_0^{2\pi}\)(1 + cos x)⁵(1 – cos x)³ dx.
Solution:
\(\int_0^{2\pi}\)(1 + cos x)⁵(1 – cos x)³ dx . (1 + cos x)³(1 + cos x)²(1 – cos x)³

Practicing the Intermediate 2nd Year Maths 2B Textbook Solutions Inter 2nd Year Maths 2B Definite Integrals Solutions Exercise 7(b) will help students to clear their doubts quickly.

Intermediate 2nd Year Maths 2B Definite Integrals Solutions Exercise 7(b)

I. Evaluate the following definite integrals.

Question 1.
\(\int_0^a(a^2x-x^3) d x\)
Solution:

Question 2.
\(\int_2^3 \frac{2 x}{1+x^2} d x\)
Solution:
I = [ln|1 + x²|]³₂
= ln 10 – ln 5
= ln(10/5)
= ln 2

Question 3.
\(\int_0^\pi \sqrt{2+2 \cos \theta} d \theta\)
Solution:

Question 4.
\(\int_0^\pi\sin^3x\cos^3xd x\)
Solution:

Question 5.
\(\int_0^2|1-x|d x\)
Solution:

Question 6.
\(\int_{-\pi / 2}^{\pi / 2} \frac{\cos x}{1+e^x} d x\)
Solution:
\(\int_{-\pi / 2}^{\pi / 2} \frac{\cos x dx}{1+e^x}\) ………….. (i)
cos x is even function
e^x is neither even nor odd.

Adding (i) and (ii) we get

Question 7.
\(\int_0^1\frac{dx}{\sqrt{3-2x}}\)
Solution:
3 – 2x = t²
-2dx = 2t dt
dx = -t dt
3 – (2.1) = t²
1 = t²
3 – 2.0 = t²

Question 8.
\(\int_0^a(\sqrt{a}-\sqrt{x})^2\)
Solution:

Question 9.
\(\int_0^{\pi / 4} \sec^4\theta d\theta\)
Solution:
Let ∫sec⁴ θ dθ = \(\int_0^{\pi / 4} \sec^2\theta(1+\tan^2\theta)d\theta\)
Put tan θ = y
sec² θ dθ = dy
θ = \(\frac{\pi}{4}\) ⇒ y= 1
θ = 0 ⇒ y = 0
I = \(\int_0^a\)(1 + y²)dy = [y + \(\frac{y^3}{3}\)]¹₀
= 1 + \(\frac{1}{3}=\frac{4}{3}\)

Question 10.
\(\int_0^3\frac{x}{\sqrt{x^2+16}}\)
Solution:

Question 11.
I = \(\int_0^1\)x.e^-x²dx
Solution:
I = \(\int_0^a\)x.e^-x²dx
⇒ -x² = t
⇒ -2x dx = dt
2x dx = -dt
x = 1 ⇒ t = 0
x = 0 ⇒ t = 1
I = \(\frac{1}{2}\)\(\int_0^a\)-e^t dt
= \(\frac{1}{2}\)[-e^t]^-1₀
= \(\frac{1}{2}\)[e⁰ – e^-1]
= \(\frac{1}{2}\)(1 – \(\frac{1}{e}\))

Question 12.
I = \(\int_1^5\frac{dx}{\sqrt{2x-1}}\)
Solution:

II. Evaluate the following integrals:

Question 1.
I = \(\int_0^4\frac{x^2}{1+x}\)
Solution:

Question 2.
\(\int_{-1}^2 \frac{x^2}{x^2+2}\)
Solution:

Question 3.
I = \(\int_0^1 \frac{x^2}{x^2+2}\)
Solution:

Question 4.
\(\int_0^{\pi / 2}\)x² sin x dx
Solution:

Question 5.
\(\int_0^4\)|2-x|dx
Solution:

Question 6.
\(\int_0^{\pi / 2}\frac{\sin^5 x}{\sin^5 x+\cos^5 x}\)
Solution:

Question 7.
\(\int_0^{\pi / 2}\frac{\sin^2 x-\cos^2 x}{\sin^3 x+\cos^3 x}\)dx
Solution:

Question 8.

Solution:

Question 9.

Solution:

Question 10.

Solution:

Question 11.

Solution:

Question 12.

Solution:

Question 13.

Solution:

Question 14.

Solution:

Question 15.

Solution:

III. Evaluate the following integrals :

Question 1.
\(\int_0^{\pi / 2}\frac{dx}{4+5\cos x}\)
Solution:

Question 2.
\(\int_a^b\sqrt{(x-a)(b-x)}\)dx
Solution:

Question 3.

Solution:

Question 4.

Solution:

Question 5.

Solution:

Question 6.
\(\int_0^a\)x(a – x)ⁿdx
Solution:
I = \(\int_0^a\)x(a – x)ⁿdx ………… (i)
\(\int_0^a\)f(x) dx = \(\int_0^a\)f(a – x)dx
I = \(\int_0^a\)(a – x).(x)ⁿdx ………… (ii)
I = \(\int_0^a\)axⁿ dx – xⁿ⁺¹ dx

Question 7.
\(\int_0^2x\sqrt{2-x}\)dx
Solution:
I = \(\int_0^2x.\sqrt{2-x}\)dx
\(\int_0^2\)f(x)dx = \(\int_0^2\)f(a – x)dx
= \(\int_0^2\)(2 – x).√x dx
= \(\int_0^2\)((2√x – x√x)) dx

Question 8.
\(\int_0^{\pi}\)x sin³ x dx
Solution:

Question 9.
\(\int_0^{\pi}\frac{x}{1+\sin x}\)dx
Solution:
\(\int_0^{\pi}\frac{x}{1+\sin x}\)dx …………… (i)

Question 10.

Solution:

Question 11.
\(\int_0^1\frac{log(1+x)}{1+x^2}\)dx
Solution:
Put x = tan θ
dx = sec² θ dθ
x = 0 ⇒ θ = 0
x = x ⇒ θ = \(\frac{\pi}{4}\)

Question 12.
\(\int_0^{\pi}\frac{x \sin x}{1+\cos^2 x}\)dx
Solution:

Question 13.
\(\int_0^{\pi/2}\frac{\sin^2 x}{\cos x+\sin x}\)dx
Solution:

Question 14.
\(\int_0^{\pi}\frac{1}{3+2\cos x}\)dx
Solution:

Question 15.
\(\int_0^{\pi/4}\)log(1 + tan x)dx
Solution:

Question 16.
\(\int_{-1}^{3/2}\)|x sin πx| dx
Solution:
We know that |x. sin πx| = x . sin πx
where -1 ≤ x ≤ 1
and |x . sin πx| = – x.sin πx where 1 < x ≤ 3/2

Question 17.
\(\int_0^1sin^{-1}\frac{2x}{1+x^2}\)dx
Solution:
\(\int_0^1sin^{-1}\frac{2x}{1+x^2}\)dx
Put x = tan θ ⇒ dx = sec² θ dθ
x = 0 ⇒ θ = 0
x = 1 ⇒ θ π/4

Question 18.
\(\int_0^1\)x tan^-1x dx
Solution:
\(\int_0^1\)x. tan^-1x dx
Put x = tan θ ⇒ dx = sec² θ dθ
x = 0 ⇒ θ = 0;

Question 19.
\(\int_0^{\pi}\frac{x\sin x}{1+\cos^2 x}\)dx
Solution:

Question 20.
Suppose that f : R → R is continuous periodic function and T is the period of it. Let a ∈ R. Then prove that for any positive integer n,

Solution:

Practicing the Intermediate 2nd Year Maths 2B Textbook Solutions Inter 2nd Year Maths 2B Definite Integrals Solutions Exercise 7(a) will help students to clear their doubts quickly.

Intermediate 2nd Year Maths 2B Definite Integrals Solutions Exercise 7(a)

I. Evaluate the following integrals as the limit of a sum.

Question 1.
\(\int_0^5(x+1) d x\)
Solution:

Question 2.
\(\int_0^4(x^2+1) d x\)
Solution:

II. Evaluate the following integrals as limit of a sum.

Question 1.
\(\int_0^4(x+e^{2x}) d x\)
Solution:

Question 2.
\(\int_0^1(x-x^2) d x\)
Solution:

Practicing the Intermediate 2nd Year Maths 2B Textbook Solutions Inter 2nd Year Maths 2B Integration Solutions Exercise 6(f) will help students to clear their doubts quickly.

Intermediate 2nd Year Maths 2B Integration Solutions Exercise 6(f)

I. Evaluate the following integrals.

Question 1.
∫e^x(1 + x²) dx
Solution:
∫e^x(1 + x²) dx = ∫e^x + ∫x² . e^x dx
= e^x + (x².e^x – 2 ∫x.e^xdx)
= e^x + x².e^x – 2(x.e^x – ∫e^x dx
= e^x + x².e^x – 2x. e^x + 2e^x + C
= e^x(x² – 2x + 3) + C

Question 2.
∫x² e^-3x dx
Solution:

Question 3.
∫x³ e^ax dx
Solution:

II.

Question 1.
Show that
∫xⁿ.e^-x dx = -xⁿ e^-x + n∫x^n-1.e^-x dx
Solution:
∫xⁿ.e^-x dx = \(\frac{x^ne^{-x}}{(-1)}\) + ∫e^-x. nx^n-1 dx
= -xⁿ . e^-x + n∫x^n-1e^-x dx

Question 2.
If I_n = ∫cosⁿ x dx, than show that
I_n = \(\frac{1}{n}\) cos^n-1 x sin x + \(\frac{n-1}{n}\) I_n-2.
Solution:
I_n = ∫cosⁿ x dx = ∫cos^n-1 x. cos x dx
= cos^n-1x . sin x – ∫sin x.(n – 1)cos^n-2x (-sin x)dx
= cos^n-1x . sin x + (n – 1) ∫cos^n-2x(1 – cos² x)dx
= cos^n-1x . sin x + (n – 1)I_n-2 – (n – 1) I_n
∴ I_n(1 + n – 1) = cos^n-1x. sin x + (n – 1) I_n-2
I_n = \(\frac{\cos ^{n-1} x \sin x}{n}+\frac{n-1}{n}\)I_n-2

III.

Question 1.
Obtain reduction formula for I_n = ∫cotⁿ x dx, n being a positive integer, n ≥ 2 and deduce the value of ∫cot⁴ x dx.
Solution:
I_n = ∫cotⁿ dx = ∫cot^n-2 x . cot² x dx
= ∫cot^n-2 x . (cosec² x – 1) dx
= ∫cot^n-2 x . cosec² x dx – I_n-2
= – \(\frac{cot^{n-1}x}{n-1}\) – I_n-2
n = 4 ⇒ I₄ = –\(\frac{cot^{3}x}{3}\) – I₂
n = 2 ⇒ I₂ = – cot x – I₀ where I₀ = ∫dx = x
I₂ = – cot x – x
I₄ = –\(\frac{cot^{3}x}{3}\) – (-cot x – x) + C
= –\(\frac{cot^{3}x}{3}\) + cot x + x + C

Question 2.
Obtain the reduction formula for I_n = ∫cosecⁿ x dx, n being a positive integer, n ≥ 2 and deduce the value of ∫cosec⁵ x dx.
Solution:
I_n = ∫cosecⁿ x dx
= cosec^n-2 x(-cot x) + ∫cot x. (n – 2)cosec^n-3 x. (cot x)dx
= -cosec^n-2x. cot x + (n – 2)∫cosec^n-2x. (cosec² x – 1)dx

Question 3.
If I_{m, n} = ∫sin^m xcosⁿ xdx, then show that
for a positive integer m ≥ 2.
Solution:
I_{m, n} = ∫(sin^m x) (cosⁿ x) dx
= ∫sin^m-1 (x).(cos x)ⁿ. sin x dx
= ∫sin^m-1 (x)(cos x)ⁿ(-sin x) dx

Question 4.
Evaluate ∫sin⁵ x cos⁴ x dx
Solution:
Reduction formula
I_{m, n} = \(\frac{-\sin ^{m-1} x \cdot \cos ^{n+1} x}{m+n}+\frac{m-1}{m+n}\).I_{m-2, n}

Question 5.
If I_n = ∫(log x)ⁿ dx, then show that I_n = x(log x)ⁿ – nI_n-1, and hence find
∫(log x)⁴ dx.
Solution:
I_n = ∫(log x)ⁿ dx
= (log x)ⁿ. x – ∫x . n . (log x)^n-1 . \(\frac{1}{x}\)dx
= x.(log x)ⁿ – n∫(log x)^n-1 dx
= x(log x)ⁿ – n . I_n-1
I₄ = x(log x)⁴ – 4 . I₃
I₃ = x(log x)³ – 3 . I₂
I₂ = x(log x)² – 2 . I₁
I₁ = x log x I₀ where I₀ = ∫dx = x
I₁ = x log x – x
I₂ = (x (log x)² – 2x log x = 2x
I₃ = x(log x)³ – 3(x (log x)² – 2x log x + 2x)
= x . (log x)³ – 3x (log x)² + 6x(log x) – 6x
I₄ = x(log x)⁴ – 4[x . (log x)³ – 3x(log x)² + 6x(log x) -6x] + C
= x[(log x)⁴ – 4(log x)³ + 12(log x)² – 24(log x) + 24] + C

Practicing the Intermediate 2nd Year Maths 2B Textbook Solutions Inter 2nd Year Maths 2B Integration Solutions Exercise 6(e) will help students to clear their doubts quickly.

Intermediate 2nd Year Maths 2B Integration Solutions Exercise 6(e)

I. Evaluate the following integrals.

Question 1.
∫\(\frac{x-1}{(x-2)(x-3)}\)dx
Solution:

Question 2.
∫\(\frac{x^2}{(x+1)(x+2)^2}\)dx
Solution:
∫\(\frac{x^2}{(x+1)(x+2)^2}\) ≡ \(\frac{A}{x+1}+\frac{B}{x+2}+\frac{C}{(x+2)^2}\)
⇒ x² = A(x + 2)² + B(x + 1)(x + 2) + (x + 1) …………….. (1)
Put x = -2 in (1)
(-2)² = A(0) + B(0) + C(-2 + 1) ⇒ C = -4
Put x = -1 in (1)
(-1)² = A(-1 + 2)² + B(0) + C(0)
⇒ A = 1
Equation coeffs. of x² in (1)
1 = A + B
⇒ B = 1 – A = 1 – 1 = 0
∴ \(\frac{x^2}{(x+1)(x+2)^2}=\frac{1}{x+1}+\frac{0}{x+2}+\frac{(-4)}{(x+2)^2}\)
∴ ∫\(\frac{x^2}{(x+1)(x+2)^2}\)dx
= ∫\(\frac{1}{x+1}\)dx – 4∫\(\frac{1}{(x+2)^2}\)dx
= log|x + 1| – 4\(\frac{(-1)}{x+2}\)
= log|x + 1| + \(\frac{4}{x+2}\) + C

Question 3.
∫\(\frac{x+3}{(x-1)(x^2+1)}\)dx
Solution:
Let \(\frac{x+3}{(x-1)(x^2+1)}=\frac{A}{x-1}+\frac{Bx+C}{x^2+1}\)
⇒ (x + 3) = A(x² + 1) + (Bx + C)(x – 1) …………….. (1)
Put x = 0 in (1)
3 = A(1) + C(-1)
⇒ A – C = 3 ⇒ C = A – 3 = 2 – 3 = -1
Equation coefficient of x² in (1)
0 = A + B
⇒ B = -A = -2
∴ \(\frac{x+3}{(x-1)(x^2+1)}=\frac{+2}{(x-1)}+\frac{-2x-1}{x^2+1}\)
∫\(\frac{x+3}{(x-1)(x^2+1)}\)dx = 2∫\(\frac{1}{x-1}\)dx
-∫\(\frac{2x}{x^2+1}\)dx – ∫\(\frac{1}{x^2+1}\)dx
= 2 log |x – 1| – log |x² + 1| – tan^-1(x) + C

Question 4.

Solution:

Question 5.
∫\(\frac{dx}{(e^x+e^{2x}}\)
Solution:

Question 6.
∫\(\frac{dx}{(x+1)(x+2)}\)
Solution:

Question 7.
∫\(\frac{1}{(e^x-1}\)dx
Solution:

Question 8.
∫\(\frac{1}{(1-x)(4+x^2)}\)dx
Solution:
Let \(\frac{1}{(1-x)(4+x^2)}=\frac{A}{1-x}+\frac{Bx+C}{4+x^2}\)
⇒ 1 = A(4 + x²) + (Bx + C)(1 – x) ………….. (1)
Put x = 1 in (1)
1 = A(4 + 1) ⇒ A = \(\frac{1}{5}\)
Put x = 0 in (1)
1 = A(4) + C(1)
⇒ C = 1 – 4A = 1 – 4(\(\frac{1}{5}\)) = \(\frac{5-4}{5}\) = \(\frac{1}{5}\)
0 = A – B
⇒ B = A = \(\frac{1}{5}\)

Question 9.
∫\(\frac{2x+3}{x^3+x^2-2x}\)dx
Solution:

Put x = 0 in (1), then
3 = A(2)(-1) + B(0) + C(0)
⇒ A = –\(\frac{3}{2}\)
Put x = 1 in (1). Then
2 + 3 = A(0) + B(0) + C(1)(3)
⇒ C = \(\frac{5}{3}\)
Put x = -2 in (1). Then
2(-2) + 3 = A(0) + B(-2)(-2 – 1) + C(0)
⇒ -1 = 6B ⇒ B = \(\frac{-1}{6}\)

II. Evaluate the following integrals.

Question 1.
∫\(\frac{dx}{6x^2-5x+1}\)
Solution:

Question 2.
∫\(\frac{dx}{x(x+1)(x+2)}\)
Solution:
\(\frac{1}{x(x+1)(x+2)}\) ≡ \(\frac{A}{x}+\frac{B}{x+1}+\frac{C}{x+2}\)
⇒ 1 ≡ A(x + 1)(x + 2) + B(x)(x + 2) + C(x)(x + 1)
Put x = 0
1 = A(1)(2) + B(0) + C(0) ⇒ A = \(\frac{1}{2}\)
Put x = -1
1 = A(0) + B(0) + C(-2)(-2 + 1)
⇒ C = \(\frac{1}{2}\)

Question 3.
∫\(\frac{3x-2}{(x-1)(x+2)(x-3)}\)dx
Solution:
\(\frac{3x-2}{(x-1)(x+2)(x-3)}\) ≡ \(\frac{A}{x-1}+\frac{B}{x+2}+\frac{C}{x-3}\)
⇒ 3x – 2 = A(x + 2)(x – 3) + B(x – 1)(x – 3) + C(x – 1)(x + 2)
Put x = 1
3(1) – 2 = A(1 + 2)(1 – 3) + B(0) + C(0)
⇒ A = \(\frac{-1}{6}\)
Put x = 3
3(3) – 2 = A(0) + B(0) + C(3 – 1)(3 + 2)
C = \(\frac{7}{10}\)
Put x = -2
3(-2) – 2 = A(0) + B(-2 – 1)(-2 – 3) + C(0) – 8
= 15B ⇒ B = \(\frac{-8}{15}\)

Question 4.
∫\(\frac{7x-4}{(x-1)^2(x+2)}\)dx
Solution:
\(\frac{7x-4}{(x-1)^2(x+2)}\) ≡ \(\frac{A}{x-1}+\frac{B}{(x-1)^2}+\frac{C}{x+2}\)
⇒ 7x – 4 = A(x – 1)(x + 2) + B(x + 2) + C(x – 1)² ………….. (1)
Put x = 1 in (1)
7 – 4 = A(0) + B(1 + 2) ⇒ B = 1
Put x = -2 in (1)
7(-2) – 4 = A(0) + B(0) + C(-2 – 1)²
⇒ -18 = 9C ⇒ C = -2
Equating coeffs. of x² in (1)
0 = A + C ⇒ A = -C = 2

III. Evaluate the following integrals.

Question 1.
∫\(\frac{1}{(x-a)(x-b)(x-c)}\)dx
Solution:

Question 2.
∫\(\frac{2x+3}{(x+3)(x^2+4)}\)dx
Solution:
\(\frac{2x+3}{(x+3)(x^2+4)}\) = \(\frac{A}{x+3}+\frac{Bx+C}{x^2+4}\)
2x + 3 = A(x² + 4) + (Bx + C)(x + 3)
x = -3 ⇒ -3 = A(9 + 4) = 13A
A = –\(\frac{3}{13}\)
Equating the coefficient of x²
0 = A + B ⇒ B = -A = \(\frac{3}{13}\)
Equating the constants
3 = 4A + 3C

Question 3.
∫\(\frac{2x^2+x+1}{(x+3)(x-2)^2}\)dx
Solution:
Let \(\frac{2x^2+x+1}{(x+3)(x-2)^2}\) = \(\frac{A}{x+3}+\frac{B}{x-2}+\frac{C}{(x-2)^2}\)
2x² + x + 1 = A(x – 2)² + B(x + 3)(x – 2) + C(x + 3)
x = 2 ⇒ 8 + 2 + 1 = C(2 + 3) = 5C
⇒ C = \(\frac{11}{5}\)
x = -3 ⇒ 18 – 3 + 1
= A(-5)² = 25 A ⇒ A = \(\frac{16}{25}\)
Equating the coefficients of x²
2 = A + B ⇒ B = 2 – A = 2 – \(\frac{16}{25}\) = \(\frac{34}{25}\)

Question 4.
∫\(\frac{dx}{x^3+1}\)dx
Solution:
\(\frac{1}{x^3+1}\) = \(\frac{1}{(x+1)(x^2-x+1)}\)
Let \(\frac{1}{x^3+1}\) = \(\frac{1}{x+1}+\frac{1}{x^2-x+1}\)
⇒ 1 = A(x² – x + 1) + (Bx + C)(x + 1) ……………. (1)
Put x = -1 in (1)
1 = A(1 + 1 + 1) + (-B + C)(0)
⇒ 3A = 1 ⇒ A = \(\frac{1}{3}\)
Put x = 0 in (1)
1 = A(1) + C(1)
⇒ C = 1 – A = 1 – \(\frac{1}{3}=\frac{2}{3}\)
Equating the coefficients of x²
O = A + B ⇒ B = -A = –\(\frac{1}{3}\)

Question 5.
∫\(\frac{\sin x \cos x}{\cos^2 x+3cos x+2}\)dx
Solution:
Put cos x = t ⇒ – sin x dx = dt
∫\(\frac{\sin x \cos x}{\cos^2 x+3cos x+2}\)dx = ∫\(\frac{-t dt}{t^2+3t+2}\)
= -∫\(\frac{t}{t^2+3t+2}\)dt …………. (1)
Let \(\frac{t}{t^2+3t+2}\) = \(\frac{t}{(t+1)(t+2)}\)
= \(\frac{A}{t+1}+\frac{B}{t+2}\)
⇒ t = A(t + 2) + B(t + 1) ………… (2)
Put t = -1 in (2)
-1 = A(-1 + 2) ⇒ A = -1
Put t = -2 in (2)
-2 = B(-2 + 1) ⇒ B = 2
∴ \(\frac{t}{t^2+3t+2}\) = \(\frac{-1}{t+1}+\frac{2}{t+2}\) ……….. (3)
∴ From (1) & (2)
∫\(\frac{\sin x.\cos x}{\cos^2 x+3cos x+2}\)dx
= -[∫\(\frac{-1}{t+1}\)dt+2∫\(\frac{1}{t+2}\)]
= ∫\(\frac{1}{t+1}\) – 2∫\(\frac{1}{t+2}\)
= log|t + 1| – 2log|t + 2| + C
= log|1 + cos x| – 2log|2 + cos x| + C
= log|1 + cos x| – log(2 + cos x)² + C
= log|\(\frac{1+\cos x}{(2+\cos x)^2}\)| + C

Practicing the Intermediate 2nd Year Maths 2B Textbook Solutions Inter 2nd Year Maths 2B Integration Solutions Exercise 6(d) will help students to clear their doubts quickly.

Intermediate 2nd Year Maths 2B Integration Solutions Exercise 6(d)

I. Evaluate the following integrals.

Question 1.
∫\(\frac{dx}{\sqrt{2x-3x^2+1}}\)
Solution:

Question 2.
∫\(\frac{\sin \theta}{\sqrt{2-\cos^2 \theta}}\)dθ
Solution:

Question 3.
∫\(\frac{\cos x}{\sin^2 x+4sin x+5}\)dx
Solution:
t = sin x ⇒ dt = cos x dx
I = ∫\(\frac{dt}{t^2+4t+5}\) = ∫\(\frac{dt}{(t+2)^2+1}\)
= tan^-1(t + 2) + C
= tan^-1(sin x + 2) + C

Question 4.
∫\(\frac{dx}{1+\cos^2 x}\)
Solution:

Question 5.
∫\(\frac{dx}{2\sin^2 x+3\cos^2 x}\)
Solution:

Question 6.
∫\(\frac{1}{1+\tan x}\)dx
Solution:

Question 7.
∫\(\frac{1}{1-\cot x}\)dx
Solution:

II. Evaluate the following integrals.

Question 1.
∫\(\sqrt{1+3x-x^2}\)dx
Solution:

Question 2.
∫\(\frac{9\cos x-\sin x}{4\sin x+5\cos x}\)dx
Solution:

Question 3.
∫\(\frac{2\cos x+3\sin x}{4\cos x+5\sin x}\)dx
Solution:
Let 2 cos c + 3 sin x = A(4 cos x + 5 sin x) + B(-4 sin x + 5 cos x)
Equating the co-efficient of sin x and cos x,
we get
4A + 5B = 2
5A – 4B = 3

Question 4.
∫\(\frac{1}{1+\sin x+\cos x}\)dx
Solution:

Question 5.
∫\(\frac{1}{3x^2+x+1}\)dx
Solution:

Question 6.
∫\(\frac{dx}{\sqrt{5-2x^2+4x}}\)
Solution:

III. Evaluate the following integrals.

Question 1.
∫\(\frac{x+1}{\sqrt{x^2-x+1}}\)
Solution:

Question 2.
∫(6x + 5)\(\sqrt{6-2x^2+x}\)dx
Solution:
Let 6x + 5 = A(1 – 4x) + B
Equating the constants
A + B = 5
B = 5 – A = 5 + \(\frac{3}{2}\) = \(\frac{13}{2}\)
∫(6x + 5)\(\sqrt{6-2x^2+x}\)dx

Question 3.
∫\(\frac{dx}{4+5\sin x}\)
Solution:
t = tan \(\frac{x}{2}\) ⇒ dt = sec² \(\frac{x}{2}\) . \(\frac{1}{2}\)dx

Question 4.
∫\(\frac{1}{2-3\cos 2x}\)dx
Solution:
t = tan x ⇒ dt = sec² x dx
= (1 + tan² x)dx
= (1 +t²)dx
dx = \(\frac{dt}{1+t^2}\)

Question 5.
∫x\(\sqrt{1+x-x^2}\)dx
Solution:
Let x = A(1 – 2x) + B
Equating the coefficients of x
1 = -2 A ⇒ A = –\(\frac{1}{2}\)
Equating the constants
0 = A + B ⇒ B = -A = \(\frac{1}{2}\)
∫x\(\sqrt{1+x-x^2}\)dx
= –\(\frac{1}{2}\)∫(1 – 2x)\(\sqrt{1+x-x^2}\)dx + \(\frac{1}{2}\)∫\(\sqrt{1+x-x^2}\)dx

Question 6.
∫\(\frac{dx}{(1+x)\sqrt{3+2x-x^2}}\)
Solution:

Question 7.
∫\(\frac{dx}{4\cos x+3\sin x}\)
Solution:
Let t = tan\(\frac{x}{2}\) so that dx = \(\frac{2dt}{1+t^2}\)

Question 8.
∫\(\frac{1}{\sin x+\sqrt{3} \cos x}\)dx
Solution:
Let t = tan \(\frac{x}{2}\) so that dx = \(\frac{2dt}{1+t^2}\)
sin x = \(\frac{2t}{1+t^2}\), cos x = \(\frac{1-t^2}{1+t^2}\)

Question 9.
∫\(\frac{dx}{5+4\cos 2x}\).
Solution:

Question 10.
∫\(\frac{2\sin x+3\cos x+4}{3\sin x+4\cos x+5}\)dx.
Solution:
Let 2 sin x + 3 cos x + 4
= A(3 sin x + 4 cos x + 5) + 3(3 cos x – 4 sin x) + C
Equating the co-efficient of
sin x, we get 3A – 4B = 2
cos x, we get 4A + 3B = 3

Equating the constants
4 = 5A + C
C = 4 – 5A = 4 – 5.\(\frac{18}{25}\) = \(\frac{2}{5}\)

Substituting in (1)
I = \(\frac{18}{25}\). x + \(\frac{1}{25}\) log|3 sin x + 4 cos x + 5| – \(\frac{4}{5\left(3+\tan \frac{x}{2}\right)}\) + C

Question 11.
∫\(\sqrt{\frac{5-x}{x-2}}\) dx on (2, 5).
Solution:

Let 5 – x = A. \(\frac{d}{dx}\)(7x – 10 – x²) + B
⇒ 5 – x = A(7 – 2x) + B
Equating coffs. of like terms
-2A = -1 ⇒ A = \(\frac{1}{2}\)
7A + B = 5
7(+\(\frac{1}{2}\)) + B = 5 ⇒ B = 5 – \(\frac{7}{2}\) = \(\frac{3}{2}\)
∴ 5 – x = \(\frac{1}{2}\)(7 – 2x) + \(\frac{3}{2}\)

Question 12.
∫\(\sqrt{\frac{1+x}{1-x}}\) dx on (-1, 1).
Solution:

Question 13.
∫\(\frac{dx}{(1 – x)\sqrt{3-2x+x^2}}\) on (-1, 3).
Solution:
Put 1 – x = \(\frac{1}{1}\) ⇒ 1 – \(\frac{1}{1}\) = x\(\frac{1}{1-x}\) = t
dx = \(\frac{1}{t^2}\)dt
3 – 2x – x² = 3 – 2(\(\frac{1-1}{1}\)) – (\(\frac{1-1}{1}\))²

Question 14.
∫\(\frac{dx}{(x + 2)\sqrt{x+1}}\) on (-1, ∞).
Solution:
Put = x + 1 = t² ⇒ dx = 2t dt and
x + 2 = 1 + t²

Question 15.
∫\(\frac{dx}{(2x + 3)\sqrt{x+2}}\) on I ⊂ (-2, ∞)\{\(\frac{-3}{2}\)}.
Solution:
Put x + 2 = t² ⇒ dx = 2t dt and
2x + 3 = 2(t² – 2) + 3 = 2t² – 1

Question 16.
∫\(\frac{1}{(1+\sqrt{x}) \sqrt{x-x^2}}\)dx on (0, 1).
Solution:

Question 17.
∫\(\frac{dx}{(x + 1)\sqrt{2x^2+3x+1}}\) on
Solution:

Question 18.
∫\(\sqrt{e^x-4}\) dx on [log_e 4, ∞]
Solution:
Put e^x – 4 = t² ⇒ e^x dx = 2t dt

Question 19.
∫\(\sqrt{1+\sec x}\) dx on [(2n – \(\frac{1}{2}\))π – (2n + \(\frac{1}{2}\))π], (n ∈ Z).
Solution:

Question 20.
∫\(\frac{dx}{1+x^4}\) on R.
Solution:

Practicing the Intermediate 2nd Year Maths 2B Textbook Solutions Inter 2nd Year Maths 2B Integration Solutions Exercise 6(c) will help students to clear their doubts quickly.

Intermediate 2nd Year Maths 2B Integration Solutions Exercise 6(c)

I. Evaluate the following integrals.

Question 1.
∫x sec² x dx on I ⊂ R\{\(\frac{(2n+1)\pi}{2}\) : n is integer}.
Solution:
∫uvdx = u∫vdx – ∫[\(\frac{d}{du}\)(U).∫vdx]dx
Let v = sec² x and u = x then
∫x sec² x dx = x(tan x) – ∫tan x dx
= x tan x – log|sec x| + C

Question 2.
∫e^x(tan^-1 x + \(\frac{1}{1+x^2}\))dx, x ∈ R.
Solution:
∫e^x[f(x) + f'(x)] dx = e^x. f(x) + C
Let f(x) = tan^-1 x so that f'(x) = \(\frac{1}{1+x^2}\)
∴ ∫e^x(tan^-1 x + \(\frac{1}{1+x^2}\))dx = e^xtan^-1 x + C

Question 3.
∫\(\frac{log x}{x^2}\) dx on (0, ∞).
Solution:
∫\(\frac{log x}{x^2}\) dx = (log x)(-\(\frac{1}{x}\)) + ∫\(\frac{1}{x}\) . \(\frac{1}{x}\)dx
= – \(\frac{1}{x}\)log x – \(\frac{1}{x}\) + C

Question 4.
∫(log x)² dx (0, ∞).
Solution:
∫(log x)² dx = (log x)² x – ∫x . 2log x.\(\frac{1}{x}\) dx
= x (log x)² – 2 ∫log x dx
= x (log x)² – 2(x . log x – ∫x\(\frac{1}{x}\)dx)
= x(log x)² – 2x. log x + 2x + c

Question 5.
∫e^x(sec x + sec x . tan x)dx on I ⊂ R\{(2n + 1)\(\frac{\pi}{2}\) : n ∈ Z}.
Solution:
∫e^x(sec x + sec x . tan x)dx = e^x.sec x + C
[∫e^x[f(x) + f'(x)]dx = e^x f(x) + C]

Question 6.
∫e^x cos x dx on R.
Solution:
I = ∫e^x cos x dx = e^x sin x – ∫sin x . e^x dx
= e^x. sin x + e^x. cos x – ∫e^x . cos x dx
= e^x(sin x + cos x) – I
2I = e^x(sin x + cos x)
I = \(\frac{e^x}{2}\)(sin x + cos x) + C

Question 7.
∫e^x(sin x + cos x) dx on R.
Solution:
∫e^x(sin x + cos x) dx
f(x) = sin x ⇒ f'(x) = cos x
∴ ∫e^x(sin x + cos x) dx = e^x. sin x + C

Question 8.
∫(tan x + log sec x)e^x dx on ((2n – \(\frac{1}{2}\))π, (2n + \(\frac{1}{2}\))π) n ∈ Z.
Solution:
t = log|sec| ⇒ dt = \(\frac{1}{\sec x}\). sec x .tan x dx
= tan x dx
∫(tan x + log sec x)e^x dx = e^x.log|sec x| + C

II. Evaluate the following integrals.

Question 1.
∫xⁿlog x dx on (0, ∞), n is a real nember and n ≠ -1.
Solution:
∫xⁿlog x dx

Question 2.
∫log (1 + x²) dx on R.
Solution:
∫log (1 + x²) dx on
= [log (1 + x²). x – ∫x\(\frac{1}{1+x^2}\)2x dx
= x log (1 + x²) – 2∫\(\frac{1+x^2-1}{1+x^2}\)dx
= x log (1 + x²) – 2∫dx + 2∫\(\frac{dx}{1+x^2}\)
= x log (1 + x²) – 2x + 2 tan^-1 x + C

Question 3.
∫\(\sqrt{x}\) log x dx on (0, ∞).
Solution:
∫\(\sqrt{x}\) log x dx

Question 4.
∫e^√x dx on (0, ∞).
Solution:
t = √x ⇒ x = t²
dx = 2t dt
∫e^√x dx = 2∫t e^t dt
= 2 [t e^t – ∫e^t dt
= 2 (t e^t – e^t) + C
= 2√x e^√x – 2e^√x + C

Question 5.
∫x² cos x dx on R.
Solution:
∫x² cosx dx = x²(sin x) – ∫sin x(2x dx)
= x² sin x + 2∫x(-sin x)dx
= x². sin x + 2[x cos x – ∫cos x dx]
= x² sin x + 2x cos x – 2 sin + C

Question 6.
∫x sin² x dx on R.
Solution:
∫x sin² x dx = \(\frac{1}{2}\)∫x(1 – cos 2x) dx
= \(\frac{1}{2}\)[∫x dx – ∫x cos 2x dx]
= \(\frac{1}{2}\)[\(\frac{x^2}{2}\) – {x . \(\frac{\sin 2x}{2}\) – \(\frac{1}{2}\)∫sin 2x dx}]
= \(\frac{x^2}{4}\) – \(\frac{x}{4}\)sin 2x + \(\frac{1}{4}\)∫sin 2x dx
= \(\frac{x^2}{4}\) – \(\frac{x}{4}\)sin 2x – \(\frac{1}{8}\)cos 2x + C

Question 7.
∫x cos² x dx on R.
Solution:
∫x cos² x dx = \(\frac{1}{2}\)∫x(1+ cos 2x)dx
= \(\frac{1}{2}\)[∫x dx + ∫x cos 2x dx]
= \(\frac{1}{2}\)[\(\frac{x^2}{2}\) + {x.\(\frac{\sin 2x}{2}\) – \(\frac{1}{2}\)∫sin 2x dx}]
= \(\frac{x^2}{4}\) + \(\frac{x}{4}\) sin 2x – \(\frac{1}{4}\) ∫sin 2x dx
= \(\frac{x^2}{4}\) + \(\frac{x}{4}\) sin 2x + \(\frac{1}{8}\)cos 2x + C

Question 8.
∫cos √x dx on R.
Solution:
t = x = t² ⇒ dx = 2t dt
I = 2∫t. cos t dt = 2 (t sin t – ∫sin t dt)
= 2(t sin t + cos t) + C
= 2√x sin √x + 2 cos√x + C

Question 9.
∫x sec² 2x dx on I ⊂ R\ {(2nπ + 1)\(\frac{\pi}{4}\) : n ∈ Z}.
Solution:
∫x sec² 2x dx = x\(\frac{\tan 2x}{2}\) – \(\frac{1}{2}\)∫tan 2x dx
= x\(\frac{\tan 2x}{2}\) – \(\frac{1}{2}\) . \(\frac{1}{2}\) log|sec 2x| + C
= x\(\frac{\tan 2x}{2}\) – \(\frac{1}{4}\) log|sec 2x| + C

Question 10.
∫x cot² x dx on I ⊂ R\{nπ : n ∈ Z}.
Solution:
∫x cot² x dx
= ∫x (cosec² x – 1)dx
= ∫x cosec²x dx – ∫x dx
= x (- cot x) + ∫cot x dx – \(\frac{x^2}{2}\)
= – x cot x + log|sin x| – \(\frac{x^2}{2}\) + C

Question 11.
∫e^x(tan x + sec² x)dx on I ⊂ R\{(2n + 1)\(\frac{\pi}{2}\) : n ∈ Z}.
Solution:
f(x) = tan x ⇒ f'(x) = sec² x dx
I = ∫e^x[f(x) + f'(x)] dx = e^x. f(x) + C
= e^x. tan x + C

Question 12.
∫e^x(\(\frac{1+log x}{x}\))dx on (0, ∞).
Solution:
∫e^x(\(\frac{1+log x}{x}\))dx = ∫e^x(log x + \(\frac{1}{x}\))dx
= e^x.log x + C

Question 13.
∫e^axsin bx dx on R, a, b ∈ R.
Solution:
Let I = ∫e^axsin bx dx …………. (1)

Question 14.
∫\(\frac{x.e^2}{(x+1)^2}\) dx on I ⊂ R\{-1}
Solution:

Question 15.
∫\(\frac{dx}{(x^2+a^2)^2}\), (a > 0) on R.
Solution:
Put x = tan t
Then dx = a sec² t dt

Question 16.
∫e^x log (e^2x + 5e^x + 6) dx on R.
Solution:
∫e^x log (e^2x + 5e^x + 6) dx
∵ e^2x + 5e^x + 6 = (e^x + 2)(e^x + 3)
= ∫e^x. log((e^x + 2)(e^x + 3))dx
= ∫e^x.{log(e^x + 2) + log(e^x + 3)}dx
= ∫e^xlog (e^x + 2)dx + ∫e^xlog (e^x + 3)dx
Put e^x = t ⇒ e^xdx = dt
= ∫log(t + 2)dt + ∫log(t + 3) dt
= log(t + 2)∫1dt – ∫{\(\frac{d}{dt}\)log(t + 2). ∫1dt}dt + log(t + 3)∫1dt – ∫{\(\frac{d}{dt}\)log(t + 3). ∫1dt}dt
= t log(t + 2) – ∫\(\frac{1}{t + 2}\))t dt + t log(t+ 3) – ∫\(\frac{1}{t + 3}\)). t dt
= t {log(t + 2) + log (t + 3)} – ∫\(\frac{t}{t + 2}\)dt – ∫\(\frac{t}{t + 3}\)dt
= t log (t² + 5t + 6) – ∫(\(\frac{t+2-2}{t + 2}\))dt – ∫(\(\frac{t+3-3}{t + 3}\))dt
= t. log (t² + 5t + 6) – ∫{1 – \(\frac{2}{t + 2}\)}dt – ∫{1 – \(\frac{3}{t + 3}\)}dt
= t log (t² + 5t + 6) – t + 2 log|t + 2| – t + 3 log|t + 3| + C
= t log (t² + 5t + 6) – 2t + 2 log|t + 2| + 3log|t + 3| + C
= e^x log(e^2x + 5e^x + 6) – 2e^x + 2log e^x + 2 + 3 log e^x + 3 + C

Question 17.
∫e^x.\(\frac{x+2}{(x+3)^2}\) dx on I ⊂ R\ {-3}
Solution:

Question 18.
∫cos (log x) dx on (0, ∞).
Solution:
Put log x = t
x = e^t
dx = e^t . dt
I = ∫e^t . cos t . dt
= e^t sin t – ∫sin.e^t dt
= e^t . sin t + cos t . e^t – ∫e^t . cos t dt
2I = e^t . (sin t + cos t)
I = \(\frac{e^t}{2}\)(sin t + cos t)
= \(\frac{x}{2}\)[sin(log x) + cos(log x)] + C

III. Evaluate the following integrals.

Question 1.
∫x tan^-1 x dx, x ∈ R
Solution:
∫x tan^-1 x dx = (tan^-1 x)\(\frac{x^2}{2}\) – \(\frac{1}{2}\)∫x².\(\frac{1}{1+x^2}\)dx

Question 2.
∫x² tan^-1 x dx, x ∈ R.
Solution:

Question 3.
∫\(\frac{tan^{-1}x}{x^2}\) dx, x ∈ I ⊂ R\{0}
Solution:

Question 4.
∫x cos^-1x dx, x ∈ (-1, 1).
Solution:
∫x cos^-1x dx

Question 5.
∫x² sin^-1x dx, x ∈ (-1, 1).
Solution:
∫x² sin^-1x dx

Question 6.
∫x log(1 + x) dx, x ∈ (-1, ∞).
Solution:
∫x log(1 + x) dx

Question 7.
∫sin √x dx, on (0, ∞).
Solution:
x = t² ⇒ dx = 2t dt
∫sin √x dx = 2∫t. sin t dt
= 2(t(-cos t) + ∫cos t dt)
= -2t cos t + 2 sin t
= -2√x cos √x + 2 sin √x + C

Question 8.
∫e^ax sin (bx + c)dx, (a, b, c ∈ R, b ≠ 0) on R.
Solution:
Let I = ∫e^ax sin (bx + c)dx

Question 9.
∫a^x cos 2x dx on R (a > 0 and a ≠ 0).
Solution:

Question 10.
∫tan^-1(\(\frac{3x-x^3}{1-3x^2}\)) dx on I ⊂ R\{-\(\frac{1}{\sqrt{3}}\), \(\frac{1}{\sqrt{3}}\)}.
Solution:
Put x = tan t ⇒ dx = sec² t dt
Then
∫tan^-1(\(\frac{3x-x^3}{1-3x^2}\)) dx
= ∫tan^-1(\(\frac{3 \tan t -\tan^3 t}{1-3\tan^2 t}\)) dx
= ∫tan^-1(tan 3t).sec² t dt
= 3∫t sec² t dt
= 3[t∫sec² t dt – ∫{\(\frac{d}{dt}\)(t) ∫sec² t dt}dt]
= 3[t(tan t) – ∫(1) tan t dt]
= 3(x tan^-1 x – log\(\sqrt{1+x^2}\)) + C
= 3x[tan^-1 x – \(\frac{3}{2}\)log (1 + x²) + C
= 3x tan^-1(x) – \(\frac{3}{2}\)log (1 + x²) + C

Question 11.
∫sinh^-1 x dx on R.
Solution:

Question 12.
∫cosh^-1 x dx on (1, ∞).
Solution:

Question 13.
∫tanh^-1x dx on (-1, 1).
Solution: