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

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

I. Evaluate the following integrals.

Question 1.

∫x sec^{2}x dx on I ⊂ R – {\(\frac{(2 n+1) \pi}{2}\) : n is an integer}.

Solution:

We use the formula for integration by parts which state that

Question 2.

\(\int e^x\left(\tan ^{-1} x+\frac{1}{1+x^2}\right) d x\), x ∈ R.

Solution:

Question 3.

\(\int \frac{\log x}{x^2} d x\) on(0, ∞).

Solution:

Question 4.

∫(log x)^{2} dx on (0, ∞).

Solution:

Take (log x)^{2} = u and 1 = v.

Then by integration by parts,

∫(log x)^{2} . 1 . dx = (log x)^{2} . x – ∫2 (log x) \(\frac{1}{x}\) . x dx

= x(log x)^{2} – 2 ∫log x . 1 . dx

= x(log x)^{2} – 2[logx . x – ∫\(\frac{1}{x}\) . 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:

Let f(x) = sec x then f'(x) = sec x tan x

∴ Using ∫e^{x} [f(x) + f'(x)] dx = e^{x} f(x) + c

we have ∫e^{x} (sec x + sec x tan x) dx = e^{x} sec x + c

Question 6.

∫e^{x} cos x dx on R.

Solution:

Let I = ∫e^{x} cos x dx

and take u = e^{x} and v = cos x.

Then using integration by parts,

I = e^{x} (sin x) – ∫e^{x} sin x dx

= e^{x} (sin x) – [e^{x}(-cos x) – ∫e^{x} (-cos x) dx]

= e^{x} sin x + e^{x} cos x – ∫e^{x} cos x dx

= e^{x} (sin x + cos x) – 1

2I = e^{x} (sin x + cos x)

I = \(\frac{1}{2}\) e^{x} (sin x + cos x) + c

∴ ∫e^{x} cos x dx = \(\frac{1}{2}\) e^{x} (sin x + cos x)

Question 7.

∫e^{x} (sin x + cos x) dx on R.

Solution:

Take f(x) = sin x then f'(x) = cos x

So by using formula ∫e^{x} [f(x) + f'(x)] dx = e^{x} f(x) + c

we have ∫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:

Take f(x) = log|sec x| then f'(x) = \(\frac{1}{\sec x}\) (sec x tan x) = tan x

So by the formula ∫e^{x} [f(x) + f'(x)] dx = e^{x} f(x) + c

we have ∫(tan x + log sec x) e^{x} dx = e^{x} log|sec x| + c

II. Evaluate the following integrals.

Question 1.

∫x^{n} log x dx on (0, ∞), n is a real number and n ≠ -1.

Solution:

Take u = log x and v = x^{n}

applying integration by parts,

Question 2.

∫log(1 + x^{2}) dx on R.

Solution:

Take log(1 + x^{2}) = u and 1 = v then

using integration by parts, we get

Question 3.

∫√x log x dx on (0, ∞).

Solution:

Take u = log x and v = x^{1/2} and

using integration by parts, we get

Question 4.

\(\int e^{\sqrt{x}} d x\) on (0, ∞).

Solution:

Question 5.

∫x^{2} cos x dx on R.

Solution:

Take x^{2} = u and cos x = v,

and using integration by parts, we get

∫x^{2} cos x dx = x^{2} (sin x) – ∫2x sin x dx

= x^{2} sin x – [2x(-cos x) – ∫2 . (-cos x) dx]

= x^{2} sin x + 2x cos x – 2∫cos x dx

= x^{2} sin x + 2x cos x – 2 sin x + c

Question 6.

∫x sin^{2}x dx on R.

Solution:

Question 7.

∫x cos^{2}x dx on R.

Solution:

Question 8.

∫cos√x dx on R.

Solution:

∫cos√x dx = ∫\(\frac{1}{\sqrt{x}}\) √x cos√x dx

Taking √x = t, we get \(\frac{1}{2 \sqrt{x}}\) dx = dt

⇒ \(\frac{\mathrm{dx}}{\sqrt{\mathrm{x}}}\) = 2 dt

∴ ∫cos √x dx = 2∫t cos t dt

using Integration by parts,

= 2[t(sin t) – ∫1 . sin t dt]

= 2[t sin t + cos t + c]

= 2[√x sin √x + cos √x ] + c

Question 9.

∫x sec^{2}2x dx on I ⊂ R \ {(2nπ + 1) \(\frac{\pi}{4}\) : n ∈ Z}

Solution:

Taking x = u and sec^{2}2x = v,

and applying integration by parts we get

∫x sec^{2}2x dx = x(\(\frac{1}{2}\) tan 2x – ∫1 . \(\frac{1}{2}\) tan 2x dx

= \(\frac{x}{2}\) tan 2x – \(\frac{1}{2}\) ∫tan 2x dx

= \(\frac{x}{2}\) tan 2x – \(\frac{1}{4}\) log|sec 2x| + c

Question 10.

∫x cot^{2}x dx on I ⊂ R \ {nπ : n ∈ Z).

Solution:

∫x cot^{2}x dx = ∫x(cosec^{2}x – 1) dx = ∫x cosec^{2}x dx – ∫x dx

Taking u = x and v = cosec^{2}x on the first integral

and using integration by parts we get

∫x cot^{2}x dx = x(-cot x) – ∫1 . (-cot x) dx – \(\frac{x^2}{2}\)

= -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 + sec^{2}x) dx on I ⊂ R \ {(2n + 1)\(\frac{\pi}{2}\) : n ∈ Z}.

Solution:

Let f(x) = tan x, then f'(x) = sec^{2}x

So by the formula ∫e^{x} [f(x) + f'(x)] dx = e^{x} f(x) + c

we have ∫e^{x} (tan x + sec^{2}x) dx = e^{x} tan x + c

Question 12.

∫\(e^x\left(\frac{1+x \log x}{x}\right) d x\) on (0, ∞). (Mar. ’13)

Solution:

Question 13.

∫e^{ax} sin bx dx on R, a, b ∈ R.

Solution:

Let I = ∫e^{ax} sin bx dx

Taking u = e^{ax} and v = sin bx

and applying integration by parts

Question 14.

\(\int \frac{x e^x}{(x+1)^2} d x\) on I ⊂ R \ {-1}.

Solution:

Question 15.

\(\int \frac{d x}{\left(x^2+a^2\right)^2}\), (a > 0) on R.

Solution:

Take substitution x = a tan θ

so that dx = a sec^{2}θ dθ

Question 16.

∫e^{x} log(e^{2x} + 5e^{x} + 6) dx on R.

Solution:

Question 17.

\(\int e^x \frac{(x+2)}{(x+3)^2} d x\) on I ⊂ R \ {-3}.

Solution:

Question 18.

∫cos(log x) dx on (0, ∞).

Solution:

Let I = ∫cos (log x) dx = ∫cos (log x) . 1 dx

Take u = cos (log x) and v = 1

and using integration by parts successively.

I = cos (log x) . x – ∫-sin (log x) \(\frac{1}{x}\) . x . dx

= x cos (log x) + ∫sin(log x) dx

= x cos (log x) + sin (log x) . x – ∫cos (log x) \(\frac{1}{x}\) . x . dx

= x cos (log x) + x . sin(log x) – ∫cos (log x) dx

= x [cos (log x) + sin (log x)] – ∫cos (log x) dx

= x [cos (log x) + sin (log x)] – I

∴ 2I = x [cos (log x) + sin (log x)]

⇒ I = \(\frac{x}{2}\) [cos (log x) + sin (log x)] + c

∴ ∫cos (log x) dx = \(\frac{x}{2}\) [cos (log x) + sin (log x)] + c

III. Evaluate the following integrals.

Question 1.

∫x tan^{-1}x dx, x ∈ R.

Solution:

Let u = tan^{-1}x and v = x then using integration by parts

Question 2.

∫x^{2} tan^{-1}x dx, x ∈ R.

Solution:

Take u = tan^{-1}x and v = x^{2}

and apply integration by parts we get

Question 3.

\(\int \frac{\tan ^{-1} x}{x^2} d x\), x ∈ I ⊂ R \ {0}.

Solution:

Question 4.

∫x cos^{-1}x dx, x ∈ (-1, 1).

Solution:

Let cos^{-1}x = θ then cos θ = x

⇒ dx = -sin θ dθ

∴ ∫x cos^{-1}x dx = ∫θ cos θ (-sin θ dθ) – \(\frac{1}{2}\) ∫θ sin 2θ dθ

Using integration by parts by taking u = θ, v = sin 2θ we get

Question 5.

∫x^{2} sin^{-1}x dx, x ∈ (-1, 1).

Solution:

Let sin^{-1}x = θ then sin θ = x

⇒ dx = cos θ dθ

∴ ∫x^{2} sin^{-1}x dx = ∫θ sin^{2}θ cos θ dθ

Using integration by parts

by choosing functions u = θ and v = sin^{2}θ cos θ, we get

Question 6.

∫x log(1 + x) dx, x ∈ (-1, ∞).

Solution:

Take u = log(1 + x) and v = x

and apply integration by parts

Question 7.

∫sin√x dx on (0, ∞).

Solution:

\(\int \frac{\sqrt{x}}{\sqrt{x}} \sin \sqrt{x} d x\)

Let √x = t then \(\frac{1}{2 \sqrt{x}}\) dx = dt

⇒ \(\frac{d x}{\sqrt{x}}\) = 2 dt

∴ ∫sin √x dx = 2∫t sin t dt,

using Integration by parts by taking u = t and v = sin t, we get

= 2[t(-cos t) – ∫1 . (-cos t) dt]

= 2[-t cos t + ∫cos t dt]

= 2[-t cos t + sin t] + c

= 2[sin √x – √x cos √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, taking u = e^{ax} and v = sin (bx + c)

and applying integration by parts,

Question 9.

∫a^{x} cos 2x dx on R (a > 0 and a ≠ 1).

Solution:

Let I = ∫a^{x} cos 2x dx

using integration by parts by taking cos 2x = u and a^{x} = v, we have

Question 10.

\(\int \tan ^{-1}\left(\frac{3 x-x^3}{1-3 x^2}\right) d x\) on I ⊂ R – \(\left\{-\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right\}\).

Solution:

Question 11.

∫sinh^{-1}x dx on R.

Solution:

Take sinh^{-1}x = u and v = 1,

Applying integration by parts we get

Question 12.

∫cosh^{-1}x dx on [1, ∞).

Solution:

∫cosh^{-1}x dx = ∫cosh^{-1}x . 1 . dx

Take u = cosh^{-1}x and v = 1 then

Question 13.

∫tanh^{-1}x dx on (-1, 1).

Solution:

∫tanh^{-1}x dx = ∫tanh^{-1}x . 1 . dx

Take u = tanh^{-1}x and v = 1

and apply integration by parts we get