NCERT Solutions for Exercise 7.11 Class 12 Maths Chapter 7 - Integrals

Integrals Class 12 Chapter 7 Exercise: 7.11

Question:1 By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

${\int }_0^{\frac{\pi }{2}}{\cos }^{2}xdx$

Answer:

We have _{$I={\int }_0^{\frac{\pi }{2}}{\cos }^{2}xdx$ .............................................................} (i)

By using

${\int }_0^{a}f(x)dx={\int }_0^{a}f(a-x)dx$

We get :-

_{$I={\int }_0^{\frac{\pi }{2}}{\cos }^{2}xdx={\int }_0^{\frac{\pi }{2}}{\cos }^2(\frac{\pi }{2}-x)dx$}

or

$I={\int }_0^{\frac{\pi }{2}}{\sin }^{2}xdx$ ................................................................ (ii)

Adding both (i) and (ii), we get :-

${\int }_0^{\frac{\pi }{2}}{\cos }^{2}xdx$ $+{\int }_0^{\frac{\pi }{2}}{\sin }^{2}xdx=2I$

or ${\int }_0^{\frac{\pi }{2}}(co{s}^{2}x+si{n}^{2}x)dx=2I$

or ${\int }_0^{\frac{\pi }{2}}1.dx=2I$

or $2I={\left[x\right]}_0^{\frac{\mathrm{\Pi }}{2}}=\frac{\mathrm{\Pi }}{2}$

or $I=\frac{\mathrm{\Pi }}{4}$

Question:2 By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

. ${\int }_0^{\frac{\pi }{2}}\frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}dx$

Answer:

We have _{$I={\int }_0^{\frac{\pi }{2}}\frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}dx$ ..........................................................................} (i)

By using ,

${\int }_0^{a}f(x)dx={\int }_0^{a}f(a-x)dx$

We get,

_{$I={\int }_0^{\frac{\pi }{2}}\frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}dx={\int }_0^{\frac{\pi }{2}}\frac{\sqrt{\sin (\frac{\pi }{2}-x)}}{\sqrt{\sin (\frac{\pi }{2}-x)}+\sqrt{\cos (\frac{\pi }{2}-x)}}dx$}

or $I={\int }_0^{\frac{\pi }{2}}\frac{\sqrt{\cos x}}{\sqrt{\cos x}+\sqrt{\sin x}}dx$ .......................................................(ii)

Adding (i) and (ii), we get,

$2I={\int }_0^{\frac{\pi }{2}}\frac{\sqrt{\sin x}+\sqrt{\cos x}}{\sqrt{\sin x}+\sqrt{\cos x}}dx$

or $2I={\int }_0^{\frac{\pi }{2}}1.dx$

or $2I={\left[x\right]}_0^{\frac{\mathrm{\Pi }}{2}}=\frac{\mathrm{\Pi }}{2}$

Thus $I=\frac{\mathrm{\Pi }}{4}$

Question:​​​​​​​3 By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

${\int }_0^{\frac{\pi }{2}}\frac{{\sin }^{\frac{3}{2}}xdx}{{\sin }^{\frac{3}{2}}x+{\cos }^{\frac{3}{2}}x}$

Answer:

We have $I={\int }_0^{\frac{\pi }{2}}\frac{{\sin }^{\frac{3}{2}}xdx}{{\sin }^{\frac{3}{2}}x+{\cos }^{\frac{3}{2}}x}$ ..................................................................(i)

By using :

${\int }_0^{a}f(x)dx={\int }_0^{a}f(a-x)dx$

We get,

$I={\int }_0^{\frac{\pi }{2}}\frac{{\sin }^{\frac{3}{2}}(\frac{\pi }{2}-x)dx}{{\sin }^{\frac{3}{2}}(\frac{\pi }{2}-x)+{\cos }^{\frac{3}{2}}(\frac{\pi }{2}-x)}$

or $I={\int }_0^{\frac{\pi }{2}}\frac{{\cos }^{\frac{3}{2}}xdx}{{\sin }^{\frac{3}{2}}x+{\cos }^{\frac{3}{2}}x}$ . ............................................................(ii)

Adding (i) and (ii), we get :

$2I={\int }_0^{\frac{\pi }{2}}\frac{(si{n}^{\frac{3}{2}}x+co{s}^{\frac{3}{2}}x)dx}{{\sin }^{\frac{3}{2}}x+{\cos }^{\frac{3}{2}}x}$

or $2I={\int }_0^{\frac{\pi }{2}}1.dx$

or $2I={\left[x\right]}_0^{\frac{\pi }{2}}=\frac{\pi }{2}$

Thus $I=\frac{\pi }{4}$

Question:4 By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

. ${\int }_0^{\frac{\pi }{2}}\frac{{\cos }^{5}xdx}{{\sin }^{5}x+{\cos }^{5}x}$

Answer:

We have $I={\int }_0^{\frac{\pi }{2}}\frac{{\cos }^{5}xdx}{{\sin }^{5}x+{\cos }^{5}x}$ ..................................................................(i)

By using :

${\int }_0^{a}f(x)dx={\int }_0^{a}f(a-x)dx$

We get,

$I={\int }_0^{\frac{\pi }{2}}\frac{{\cos }^5(\frac{\pi }{2}-x)dx}{{\sin }^5(\frac{\pi }{2}-x)+{\cos }^5(\frac{\pi }{2}-x)}$

or $I={\int }_0^{\frac{\pi }{2}}\frac{{\sin }^{5}xdx}{{\sin }^{5}x+{\cos }^{5}x}$ . ............................................................(ii)

Adding (i) and (ii), we get :

$2I={\int }_0^{\frac{\pi }{2}}\frac{(si{n}^{5}x+co{s}^{5}x)dx}{{\sin }^{5}x+{\cos }^{5}x}$

or $2I={\int }_0^{\frac{\pi }{2}}1.dx$

or $2I={\left[x\right]}_0^{\frac{\pi }{2}}=\frac{\pi }{2}$

Thus $I=\frac{\pi }{4}$

Question:5 By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

${\int }_{-5}^5|x+2|dx$

Answer:

We have, $I={\int }_{-5}^5|x+2|dx$

For opening the modulas we need to define the bracket :

If (x + 2) < 0 then x belongs to (-5, -2). And if (x + 2) > 0 then x belongs to (-2, 5).

So the integral becomes :-

$I={\int }_{-5}^{-2}-(x+2)dx+{\int }_{-2}^5(x+2)dx$

or $I=-{[\frac{x^2}{2}+2x]}_{-5}^{-2}+{[\frac{x^2}{2}+2x]}_{-2}^5$

This gives $I=29$

Question:6​​​​​​​ By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

${\int }_2^8|x-5|dx$

Answer:

We have, $I={\int }_2^8|x-5|dx$

For opening the modulas we need to define the bracket :

If (x - 5) < 0 then x belongs to (2, 5). And if (x - 5) > 0 then x belongs to (5, 8).

So the integral becomes:-

$I={\int }_2^5-(x-5)dx+{\int }_5^8(x-5)dx$

or $I=-{[\frac{x^2}{2}-5x]}_2^5+{[\frac{x^2}{2}-5x]}_5^8$

This gives $I=9$

Question:7 By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

${\int }_0^{1}x(1-x{)}^{n}dx$

Answer:

We have $I={\int }_0^{1}x(1-x{)}^{n}dx$

U sing the property : -

${\int }_0^{a}f(x)dx={\int }_0^{a}f(a-x)dx$

We get : -

$I={\int }_0^{1}x(1-x{)}^{n}dx={\int }_0^1(1-x)(1-(1-x){)}^{n}dx$

or $I={\int }_0^1(1-x)x^{n}dx$

or $I={\int }_0^1(x^n-x^{n+1})dx$

or $={[\frac{x^{n+1}}{n+1}-\frac{x^{n+2}}{n+2}]}_0^1$

or $=[\frac{1}{n+1}-\frac{1}{n+2}]$

or $I=\frac{1}{(n+1)(n+2)}$

Question:8 By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

${\int }_0^{\frac{\pi }{4}}\log (1+\tan x)dx$

Answer:

We have $I={\int }_0^{\frac{\pi }{4}}\log (1+\tan x)dx$

By using the identity

${\int }_0^{a}f(x)dx={\int }_0^{a}f(a-x)dx$

We get,

$I={\int }_0^{\frac{\pi }{4}}\log (1+\tan x)dx={\int }_0^{\frac{\pi }{4}}\log (1+\tan (\frac{\pi }{4}-x))dx$

or $I={\int }_0^{\frac{\pi }{4}}\log (1+\frac{1-\tan x}{1+\tan x})dx$

or $I={\int }_0^{\frac{\pi }{4}}\log (\frac{2}{1+\tan x})dx$

or $I={\int }_0^{\frac{\pi }{4}}\log 2dx-{\int }_0^{\frac{\pi }{4}}\log (1+\tan x)dx$

or $I={\int }_0^{\frac{\pi }{4}}\log 2dx-I$

or $2I={[x\log 2]}_0^{\frac{\mathrm{\Pi }}{4}}$

or $I=\frac{\mathrm{\Pi }}{8}\log 2$

Question:9 By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

${\int }_0^{2}x\sqrt{2-x}dx$

Answer:

We have $I={\int }_0^{2}x\sqrt{2-x}dx$

By using the identity

${\int }_0^{a}f(x)dx={\int }_0^{a}f(a-x)dx$

We get :

$I={\int }_0^{2}x\sqrt{2-x}dx={\int }_0^2(2-x)\sqrt{2-(2-x)}dx$

or $I={\int }_0^2(2-x)\sqrt{x}dx$

or $I={\int }_0^2(2\sqrt{x}-x^{\frac{3}{2}}dx$

or $={[\frac{4}{3}{x}^{\frac{3}{2}}-\frac{2}{5}{x}^{\frac{5}{2}}]}_0^2$

or $=\frac{4}{3}(2{)}^{\frac{3}{2}}-\frac{2}{5}(2{)}^{\frac{5}{2}}$

or $I=\frac{16\sqrt{2}}{15}$

Question:10 By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

${\int }_0^{\frac{\pi }{2}}(2\log \sin x-\log \sin 2x)dx$

Answer:

We have $I={\int }_0^{\frac{\pi }{2}}(2\log \sin x-\log \sin 2x)dx$

or $I={\int }_0^{\frac{\pi }{2}}(2\log \sin x-\log (2\sin x\cos x))dx$

or $I={\int }_0^{\frac{\pi }{2}}(\log \sin x-\log \cos x-\log 2)dx$ ..............................................................(i)

By using the identity :

${\int }_0^{a}f(x)dx={\int }_0^{a}f(a-x)dx$

We get :

$I={\int }_0^{\frac{\pi }{2}}(\log \sin (\frac{\pi }{2}-x)-\log \cos (\frac{\pi }{2}-x)-\log 2)dx$

or $I={\int }_0^{\frac{\pi }{2}}(\log \cos x-\log \sin x-\log 2)dx$ ....................................................................(ii)

Adding (i) and (ii) we get :-

$2I={\int }_0^{\frac{\pi }{2}}(-\log 2-\log 2)dx$

or $I=-\log 2\left[\frac{\mathrm{\Pi }}{2}\right]$

or $I=\frac{\mathrm{\Pi }}{2}\log \frac{1}{2}$

Question:11 By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

${\int }_{\frac{-\pi }{2}}^{\frac{\pi }{2}}{\sin }^{2}xdx$

Answer:

We have $I={\int }_{\frac{-\pi }{2}}^{\frac{\pi }{2}}{\sin }^{2}xdx$

We know that sin ² x is an even function. i.e., sin ² (-x) = (-sinx) ² = sin ² x.

Also,

$I={\int }_{-a}^{a}f(x)dx=2{\int }_0^{a}f(x)dx$

So,

$I=2{\int }_0^{\frac{\pi }{2}}{\sin }^{2}xdx=2{\int }_0^{\frac{\pi }{2}}\frac{(1-\cos 2x)}{2}dx$

or $={[x-\frac{\sin 2x}{2}]}_0^{\frac{\mathrm{\Pi }}{2}}$

or $I=\frac{\mathrm{\Pi }}{2}$

Question:12 By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

${\int }_0^{\pi }\frac{xdx}{1+\sin x}$

Answer:

We have $I={\int }_0^{\pi }\frac{xdx}{1+\sin x}$ ..........................................................................(i)

By using the identity :-

${\int }_0^{a}f(x)dx={\int }_0^{a}f(a-x)dx$

We get,

$I={\int }_0^{\pi }\frac{xdx}{1+\sin x}={\int }_0^{\pi }\frac{(\mathrm{\Pi }-x)dx}{1+\sin (\mathrm{\Pi }-x)}$

or $I={\int }_0^{\pi }\frac{(\mathrm{\Pi }-x)dx}{1+\sin x}$ ............................................................................(ii)

Adding both (i) and (ii) we get,

$2I={\int }_0^{\pi }\frac{\mathrm{\Pi }}{1+\sin x}dx$

or $2I=\mathrm{\Pi }{\int }_0^{\pi }\frac{1-\sin x}{(1+\sin x)(1-\sin x)}dx=\mathrm{\Pi }{\int }_0^{\pi }\frac{1-\sin x}{{\cos }^{2}x}dx$

or $2I=\mathrm{\Pi }{\int }_0^{\pi }({\sec }^2-\tan x\sec x)xdx$

or $I=\mathrm{\Pi }$

Question:13 By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

${\int }_{\frac{-\pi }{2}}^{\frac{\pi }{2}}{\sin }^{7}xdx$

Answer:

We have $I={\int }_{\frac{-\pi }{2}}^{\frac{\pi }{2}}{\sin }^{7}xdx$

We know that ${\sin }^{7}x$ is an odd function.

So the following property holds here:-

${\int }_{-a}^{a}f(x)dx=0$

Hence

$I={\int }_{\frac{-\pi }{2}}^{\frac{\pi }{2}}{\sin }^{7}xdx=0$

Question:14 By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

${\int }_0^{2\pi }{\cos }^{5}xdx$

Answer:

We have $I={\int }_0^{2\pi }{\cos }^{5}xdx$

I t is known that :-

${\int }_0^{2a}f(x)dx=2{\int }_0^{a}f(x)dx$ If f (2a - x) = f(x)

$=0$ If f (2a - x) = - f(x)

Now, using the above property

${\cos }^5(\mathrm{\Pi }-x)=-{\cos }^{5}x$

Therefore, $I=0$

Question:15 By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

Answer:

We have $I={\int }_0^{\frac{\pi }{2}}\frac{\sin x-\cos x}{1+\sin x\cos x}dx$ ................................................................(i)

By using the property :-

${\int }_0^{a}f(x)dx={\int }_0^{a}f(a-x)dx$

We get ,

$I={\int }_0^{\frac{\pi }{2}}\frac{\sin (\frac{\pi }{2}-x)-\cos (\frac{\pi }{2}-x)}{1+\sin (\frac{\pi }{2}-x)\cos (\frac{\pi }{2}-x)}dx$

or $I={\int }_0^{\frac{\pi }{2}}\frac{\cos x-\sin x}{1+\sin x\cos x}dx$ ......................................................................(ii)

Adding both (i) and (ii), we get

$2I={\int }_0^{\frac{\pi }{2}}\frac{0}{1+\sin x\cos x}dx$

Thus I = 0

Question:16 By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

${\int }_0^{\pi }\log (1+\cos x)dx$

Answer:

We have $I={\int }_0^{\pi }\log (1+\tan x)dx$ .....................................................................................(i)

By using the property:-

${\int }_0^{a}f(x)dx={\int }_0^{a}f(a-x)dx$

We get,

or

$I={\int }_0^{\pi }\log (1+\cos (\mathrm{\Pi }-x))dx$

$I={\int }_0^{\pi }\log (1-\cos x)dx$ ....................................................................(ii)

Adding both (i) and (ii) we get,

$2I={\int }_0^{\pi }\log (1+\cos x)dx+{\int }_0^{\pi }\log (1-\cos x)dx$

or $2I={\int }_0^{\pi }\log (1-{\cos }^{2}x)dx={\int }_0^{\pi }\log {\sin }^{2}xdx$

or $2I=2{\int }_0^{\pi }\log \sin xdx$

or $I={\int }_0^{\pi }\log \sin xdx$ ........................................................................(iii)

or $I=2{\int }_0^{\frac{\pi }{2}}\log \sin xdx$ ........................................................................(iv)

or $I=2{\int }_0^{\frac{\pi }{2}}\log \cos xdx$ .....................................................................(v)

Adding (iv) and (v) we get,

$I=-\pi \log 2$

Question:17 By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

${\int }_0^a\frac{\sqrt{x}}{\sqrt{x}+\sqrt{a-x}}dx$

Answer:

We have $I={\int }_0^a\frac{\sqrt{x}}{\sqrt{x}+\sqrt{a-x}}dx$ ................................................................................(i)

By using, we get

${\int }_0^{a}f(x)dx={\int }_0^{a}f(a-x)dx$

We get,

$I={\int }_0^a\frac{\sqrt{x}}{\sqrt{x}+\sqrt{a-x}}dx={\int }_0^a\frac{\sqrt{(a-x)}}{\sqrt{(a-x)}+\sqrt{x}}dx$ .................................................................(ii)

Adding (i) and (ii) we get :

$2I={\int }_0^a\frac{\sqrt{x}+\sqrt{a-x}}{\sqrt{x}+\sqrt{a-x}}dx$

or $2I={\left[x\right]}_0^a=a$

or $I=\frac{a}{2}$

Question:18 By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

${\int }_0^4|x-1|dx$

Answer:

We have, $I={\int }_0^4|x-1|dx$

For opening the modulas we need to define the bracket :

If (x - 1) < 0 then x belongs to (0, 1). And if (x - 1) > 0 then x belongs to (1, 4).

So the integral becomes:-

$I={\int }_0^1-(x-1)dx+{\int }_1^4(x-1)dx$

or $I={[x-\frac{x^2}{2}]}_0^1+{[\frac{x^2}{2}-x]}_1^4$

This gives $I=5$

Question:19 Show 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 $g(x)+g(a-x)=4$

Answer:

Let $I={\int }_0^{a}f(x)g(x)dx$ ........................................................(i)

This can also be written as :

$I={\int }_0^{a}f(a-x)g(a-x)dx$

or $I={\int }_0^{a}f(x)g(a-x)dx$ ................................................................(ii)

Adding (i) and (ii), we get,

$2I={\int }_0^{a}f(x)g(a-x)dx+{\int }_0^{a}f(x)g(x)dx$

$2I={\int }_0^{a}f(x)4dx$

or $I=2{\int }_0^{a}f(x)dx$

Question:​​​​​​​20 Choose the correct answer in Exercises 20 and 21.

The value of is ${\int }_{\frac{-\pi }{2}}^{\frac{\pi }{2}}(x^3+x\cos x+{\tan }^{5}x+1)dx$ is

(A) 0

(B) 2

(C) $\pi$

(D) 1

Answer:

We have

$I={\int }_{\frac{-\pi }{2}}^{\frac{\pi }{2}}(x^3+x\cos x+{\tan }^{5}x+1)dx$

This can be written as :

$I={\int }_{\frac{-\pi }{2}}^{\frac{\pi }{2}}{x}^{3}dx+{\int }_{\frac{-\pi }{2}}^{\frac{\pi }{2}}x\cos x+{\int }_{\frac{-\pi }{2}}^{\frac{\pi }{2}}{\tan }^{5}x+{\int }_{\frac{-\pi }{2}}^{\frac{\pi }{2}}1dx$

Also if a function is even function then ${\int }_{-a}^{a}f(x)dx=2{\int }_0^{a}f(x)dx$

And if the function is an odd function then : ${\int }_{-a}^{a}f(x)dx=0$

Using the above property I become:-

$I=0+0+0+2{\int }_0^{\frac{\mathrm{\Pi }}{2}}1.dx$

or $I=2{\left[x\right]}_0^{\frac{\mathrm{\Pi }}{2}}$

or $I=\mathrm{\Pi }$

Question:​​​​​​​21 Choose the correct answer in Exercises 20 and 21.

The value of ${\int }_0^{\frac{\pi }{2}}\log \left(\frac{4+3\sin x}{4+3\cos x}\right)dx$ is

Answer:

We have

$I={\int }_0^{\frac{\pi }{2}}\log \left(\frac{4+3\sin x}{4+3\cos x}\right)dx$ .................................................................................(i)

By using :

${\int }_0^{a}f(x)dx={\int }_0^{a}f(a-x)dx$

We get,

$I={\int }_0^{\frac{\pi }{2}}\log \left(\frac{4+3\sin x}{4+3\cos x}\right)dx={\int }_0^{\frac{\pi }{2}}\log \left(\frac{4+3\sin (\frac{\pi }{2}-x)}{4+3\cos (\frac{\pi }{2}-x)}\right)dx$

or $I={\int }_0^{\frac{\pi }{2}}\log \left(\frac{4+3\cos x}{4+3\sin x}\right)dx$ .............................................................................(ii)

Adding (i) and (ii), we get:

$2I={\int }_0^{\frac{\pi }{2}}\log \left(\frac{4+3\sin x}{4+3\cos x}\right)dx+{\int }_0^{\frac{\pi }{2}}\log \left(\frac{4+3\cos x}{4+3\sin x}\right)dx$

or $2I={\int }_0^{\frac{\pi }{2}}\log 1.dx$

Thus $I=0$