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.
∫ex(1 + x²) dx
Solution:
∫ex(1 + x²) dx = ∫ex + ∫x² . ex dx
= ex + (x².ex – 2 ∫x.exdx)
= ex + x².ex – 2(x.ex – ∫ex dx
= ex + x².ex – 2x. ex + 2ex + C
= ex(x² – 2x + 3) + C

Question 2.
∫x² e-3x dx
Solution:

Question 3.
∫x³ eax dx
Solution:

II.

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

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

III.

Question 1.
Obtain reduction formula for In = ∫cotn x dx, n being a positive integer, n ≥ 2 and deduce the value of ∫cot4 x dx.
Solution:
In = ∫cotn dx = ∫cotn-2 x . cot² x dx
= ∫cotn-2 x . (cosec² x – 1) dx
= ∫cotn-2 x . cosec² x dx – In-2
= – $$\frac{cot^{n-1}x}{n-1}$$ – In-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 In = ∫cosecn x dx, n being a positive integer, n ≥ 2 and deduce the value of ∫cosec5 x dx.
Solution:
In = ∫cosecn x dx
= cosecn-2 x(-cot x) + ∫cot x. (n – 2)cosecn-3 x. (cot x)dx
= -cosecn-2x. cot x + (n – 2)∫cosecn-2x. (cosec² x – 1)dx

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

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

Question 5.
If In = ∫(log x)n dx, then show that In = x(log x)n – nIn-1, and hence find
∫(log x)4 dx.
Solution:
In = ∫(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 . In-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