Students must practice this TS Intermediate Maths 2B Solutions Chapter 6 Integration Ex 6(f) to find a better approach to solving the problems.

## TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Exercise 6(f)

I. Evaluate the following integrals.

Question 1.

∫e^{x} (1 + x^{2}) dx

Solution:

∫e^{x} (1 + x^{2}) dx = ∫(e^{x} + e^{x} x^{2}) dx

= ∫e^{x} dx + ∫e^{x} x^{2} dx

= e^{x} + x^{2} e^{x} – ∫2x e^{x} dx

= e^{x} + x^{2} e^{x} – [2x e^{x} – ∫2e^{x} dx]

= e^{x} + x^{2} e^{x} – 2x e^{x} + 2e^{x} + c

= e^{x} [1 + x^{2} – 2x + 2] + c

= e^{x} (x^{2} – 2x + 3) + c

[Integration by parts ∫uv dx = u∫v dx – ∫(\(\frac{\mathrm{d}}{\mathrm{dx}}\)(u) ∫v dx) is used]

Question 2.

∫x^{2} e^{-3x} dx

Solution:

Use integration by parts successively using u = x^{2} and v = e^{-3x} then

Question 3.

∫x^{3} e^{ax} dx

Solution:

II.

Question 1.

Show that ∫x^{n} e^{-x} dx = -x^{n} e^{-x} + n ∫x^{n-1} e^{-x} dx

Solution:

∫x^{n} e^{-x} dx using integration by parts

= x^{n} (e^{-x}) – ∫n x^{n-1} (-e^{-x}) dx

= -x^{n} e^{-x} + n∫x^{n-1} e^{-x} dx

Question 2.

If I_{n} = ∫cos^{n}x dx, then show that I_{n} = \(\frac{1}{n} \cos ^{n-1} x \sin x+\frac{n-1}{n} I_{n-2}\)

Solution:

III.

Question 1.

Obtain the reduction formula for I_{n} = ∫cot^{n}x dx, n being a positive integer, n ≥ 2 and deduce the value of ∫cot^{4}x dx. (May ’11)

Solution:

Question 2.

Obtain the reduction formula for I_{n} = ∫cosec^{n}x dx, n being a positive integer, n ≥ 2 and deduce the value of ∫cosec^{5}x dx.

Solution:

Question 3.

If I_{m,n} = ∫sin^{m}x cos^{n}x dx, then show that \(I_{m, n}=-\frac{\sin ^{m-1} x \cos ^{n+1} x}{m+n}+\frac{m-1}{m+n} I_{m-2, n}\) for a positive Integer n and an integer m ≥ 2.

Solution:

Question 4.

Evaluate ∫sin^{5}x cos^{4}x dx

Solution:

Denote I_{5,4} = ∫sin^{5}x cos^{4}x dx using the above reduction formula

Question 5.

If I_{n} = ∫(log x)^{n} dx then show that I_{n} = x(log x)^{n} – n I_{n-1} and hence find ∫(log x)^{4} dx.

Solution:

Put n = 4, 3, 2, 1 we get

I_{4} = x(log x)^{4} – 4I_{3}

= x(log x)^{4} – 4[x(log x)^{3} – 3I_{2}]

= x(log x)^{4} – 4x(log x)^{3} + 12[x(log x)^{2} – 2I_{1}]

= x(log x)^{4} – 4x(log x)^{3} + 12x(log x)^{2} – 24I_{1}]

= x(log x)^{4} – 4x(log x)^{3} + 12x(log x)^{2} – 24 ∫log x dx

= x(log x)^{4} – 4x(log x)^{3} + 12x(log x)^{2} – 24[x log x – x] + c

= x(log x)^{4} – 4x(log x)^{3} + 12x(log x)^{2} – 24x log x – 24x + c