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:
Inter 2nd Year Maths 2B Integration Solutions Ex 6(f) 1

Question 3.
∫x³ eax dx
Solution:
Inter 2nd Year Maths 2B Integration Solutions Ex 6(f) 2

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

Inter 2nd Year Maths 2B Integration Solutions Ex 6(f)

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
Inter 2nd Year Maths 2B Integration Solutions Ex 6(f) 3

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
Inter 2nd Year Maths 2B Integration Solutions Ex 6(f) 4
Inter 2nd Year Maths 2B Integration Solutions Ex 6(f) 5
Inter 2nd Year Maths 2B Integration Solutions Ex 6(f) 6

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
Inter 2nd Year Maths 2B Integration Solutions Ex 6(f) 7

 

Inter 2nd Year Maths 2B Integration Solutions Ex 6(f)

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