Gujarat Board GSEB Textbook Solutions Class 12 Maths Chapter 7 Integrals Ex 7.11 Textbook Questions and Answers.

## Gujarat Board Textbook Solutions Class 11 Maths Chapter 7 Integrals Ex 7.11

By using properties of definite integrals, evaluate the following:

Question 1.

\(\int_{0}^{\frac{\pi}{2}}\) cos^{2}x

Solution:

Question 2.

\(\frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}\)dx

Solution:

Question 3.

\(\int_{0}^{\frac{\pi}{2}} \frac{\sin ^{\frac{3}{2}} x}{\sin ^{\frac{3}{2}} x+\cos ^{\frac{3}{2}} x}\)dx

Solution:

Question 4.

\(\int_{0}^{\frac{\pi}{2}} \frac{\cos ^{5} x}{\sin ^{5} x+\cos ^{5} x}\)

Solution:

Question 5.

\(\int_{-5}^{5}\)|x + 2|dx

Solution:

Question 6.

\(\int_{2}^{8}\)|x – 5|dx

Solution:

Question 7.

\(\int_{0}^{1}\)x(1 – x)^{n}dx

Solution:

Question 8.

\(\int_{0}^{\frac{\pi}{2}}\)log(1 + tan x)dx

Solution:

Question 9.

\(\int_{0}^{2}\)x\(\sqrt{2-x}\)dx

Solution:

Question 10.

\(\int_{0}^{\frac{\pi}{2}}\)(2log sin x – log sin2x)dx

Solution:

Question 11.

\(\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin ^{2} x d x\)

Solution:

Let f(x) = sin^{2}x. Then,

f(- x) = [sin(- x)]^{2} = (- sin x)^{2} = sin^{2}x = f(x).

âˆ´ f(x) is an even function.

Question 12.

\(\int_{0}^{Ï€}\) \(\frac{xdx}{1+sinx}\)

Solution:

Adding (1) and (2), we have:

Question 13.

\(\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin ^{7} x d x\)

Solution:

Let f(x) = sin^{7}x

f(- x) = [sin (- x)]^{7} = (- sin x)^{7} = – sin^{7}x = – f(x)

â‡’ f(x) is an odd function of x.

But \(\int_{-a}^{a}\)f(x) dx = 0 when x is odd.

âˆ´ \(\int_{\frac{\pi}{2}}^{\frac{\pi}{2}} \sin ^{7} x\) dx = 0.

Question 14.

\(\int_{0}^{2Ï€}\) cos^{5}x

Solution:

f(x) = cos^{5}x

âˆ´ f(2Ï€ – x) = cos5(2Ï€ – x) = cos^{5}x = f(x).

âˆ´ I = \(\int_{0}^{2Ï€}\) cos^{5}x dx = 2\(\int_{0}^{Ï€}\)cos^{5}x dx.

Again taking g(x) = cos^{5}x, we get

g(Ï€ – x) = cos^{5}(Ï€ – x) = – cos^{5}x = – g(x)

âˆ´ I = 0. (Because g is an odd function.)

Hence, \(\int_{0}^{2Ï€}\)cos^{5}x dx = 0.

Question 15.

\(\int_{0}^{\frac{\pi}{2}}\) \(\frac{sinx-cosx}{1+sinxcosx}\)dx

Solution:

Adding (1) and (2), we get

Question 16.

\(\int_{0}^{Ï€}\)log(1 + cosx)dx

Solution:

Adding (4) and (5), we get

Put 2x = t so that 2 dx = dt.

Question 17.

\(\int_{0}^{a} \frac{\sqrt{x}}{\sqrt{x}+\sqrt{a-x}}\)dx

Solution:

Question 18.

\(\int_{0}^{1}\)|x – 1|dx

Solution:

Question 19.

Prove that \(\int_{0}^{a}\) f(x)g(x) dx = 2\(\int_{0}^{a}\) f(x)dx,

if f and g are defined as f(x) = f(a – x) and g(x) + g(x – a) = 4.

Solution:

Hence, the result.

Choose the correct answers in questions 20 and 21:

Question 20.

The value of \(\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\)(x^{3} + x cosx + tan^{5}x + 1) dx is

(A) 0

(B) 2

(C) Ï€

(D) 1

Solution:

âˆ´ Part(C) is the correct answer.

Question 21.

The value of \(\int_{0}^{\frac{\pi}{2}}\) log (\(\frac{4+3sinx}{4+3cosx}\)) dx is

(A) 2

(B) \(\frac{3}{4}\)

(C) 0

(D) – 2

Solution:

âˆ´ I = 0.

âˆ´ Part(C) is the correct answer.