Class 12 NCERT Solutions- Mathematics Part ii – Chapter 7– Integrals Exercise 7.10

Class 12 NCERT Mathematics Part ii Chapter 7 Integrals Exercise 7.10 is about using the substitution method to evaluate the given integrals. The below article provides easy solutions for the questions in the exercise.

Question 1

[Tex]\int_{0}^{1}\frac{x}{x^{2}+1}dx[/Tex]

Solution:

[Tex]\int_{0}^{1}\frac{x}{x^{2}+1} [/Tex]

Let [Tex]x^{2}+1=t \rightarrow 2x dx = dt[/Tex]

When [Tex]x=0, t=1[/Tex] and when [Tex]x=1, t=2[/Tex]

[Tex]\therefore \int_{0}^{1}\frac{x}{x^{2}+1}dx = \frac{1}{2}\int_{1}^{2}\frac{dt}{t}[/Tex]

[Tex]=\frac{1}{2} [\log|t|]_{1}^{2}[/Tex]

[Tex]=\frac{1}{2} [log2-log1][/Tex]

[Tex]=\frac{1}{2}log2[/Tex]

Question 2

[Tex]\int_{0}^{\frac{\pi}{2}}\sqrt{sin\phi } cos^{5}\phi d\phi[/Tex]

Solution:

Let [Tex]I = \int_{0}^{\frac{\pi}{2}}\sqrt{sin\phi } cos^{5}\phi d\phi = \int_{0}^{\frac{\pi}{2}}\sqrt{sin\phi } cos^{4}\phi cos\phi d\phi[/Tex]

Also, let [Tex]sin \phi=t \Rightarrow cos\phi d\phi = dt[/Tex]

When [Tex]\phi=0, t=0[/Tex] and when [Tex]\phi=\frac{\pi}{2}, t=1[/Tex]

[Tex]\therefore I=\int_{0}^{1}\sqrt{t}(1-t^2)^2dt[/Tex]

[Tex]=\int_{0}^{1}t^{\frac{1}{2}}(1+t^4-2t^2)dt[/Tex]

[Tex]=\int_{0}^{1}\left [ t^{\frac{1}{2}}+t^{\frac{9}{2}}-2t^{\frac{5}{2}} \right ]dt[/Tex]

[Tex]=\left [ \frac{t^{\frac{3}{2}}}{\frac{3}{2}}+\frac{t^{\frac{11}{2}}}{\frac{11}{2}}-\frac{t^{\frac{7}{2}}}{\frac{7}{2}} \right ]_{0}^{1}[/Tex]

[Tex]=\frac{2}{3}+\frac{2}{11}-\frac{4}{7}[/Tex]

[Tex]=\frac{154+42-132}{231}[/Tex]

[Tex]=\frac{64}{231}[/Tex]

Question 3

[Tex]\int_{0}^{1}sin^{-1}\left( \frac{2x}{1+x^{2}} \right)dx[/Tex]

Solution:

Let [Tex]I = \int_{0}^{1}sin^{-1}\left ( \frac{2x}{1+x^{2}} \right )dx[/Tex]

Also, let [Tex]x=tan\theta \Rightarrow dx = sec^{2}\theta d\theta[/Tex]

When [Tex]x=0[/Tex], [Tex]\theta=0[/Tex] and when x=1, [Tex]\theta=\frac{\pi}{4}[/Tex]

[Tex]I = \int_{0}^{\frac{\pi}{4}}sin^{-1}\left ( \frac{2tan\theta}{1+tan^{2}\theta} \right )sec^{2}\theta d\theta[/Tex]

[Tex]=\int_{0}^{\frac{\pi}{4}}sin^{-1}\left ( sin2\theta \right )sec^{2}\theta d\theta [/Tex]

[Tex]=\int_{0}^{\frac{\pi}{4}}2\theta \cdot sec^{2}\theta d\theta [/Tex]

[Tex]2\int_{0}^{\frac{\pi}{4}}\theta \cdot sec^{2}\theta d\theta [/Tex]

taking [Tex]\theta[/Tex] as first function and [Tex]sec^2\theta[/Tex] as second function and integrating by parts, we obtain

[Tex]I=2\left [ \theta\int sec^2 \theta d\theta – \int\left \{ \left ( \frac{d}{dx}\theta \right )\int sec^2\theta d\theta \right \}d\theta \right ]_{0}^{\frac{\pi}{4}} [/Tex]

[Tex]=2[\theta tan\theta – \int tan\theta d\theta]_{0}^{\frac{\pi}{4}}[/Tex]

[Tex]=2[\theta tan\theta +log|cos\theta|]_{0}^{\frac{\pi}{4}}[/Tex]

[Tex]=2\left [ \frac{\pi}{4} tan\frac{\pi}{4} +log|cos\frac{\pi}{4}|-log|cos0| \right ][/Tex]

[Tex]=2\left [ \frac{\pi}{4} + log\left ( \frac{1}{\sqrt{2}} \right ) -log1 \right ][/Tex]

[Tex]=2\left [ \frac{\pi}{4} -\frac{1}{2}log2 \right ][/Tex]

[Tex]=\frac{\pi}{2} – log2[/Tex]

Question 4

[Tex]\int_{0}^{2}x\sqrt{x+2}[/Tex] (Put x+2 = t² )

Solution:

[Tex]\int_{0}^{2}x\sqrt{x+2}dx[/Tex]

Let[Tex] x+2 = t^2 [/Tex]and [Tex]dx = 2tdt[/Tex]

When [Tex]x=0,[/Tex] [Tex]t=\sqrt{2}[/Tex] and when[Tex] x=2, t=2[/Tex]

[Tex]\therefore \int_{0}^{2}x\sqrt{x+2}dx = \int_{\sqrt{2}}^{2}\left ( t^2-2 \right )\sqrt{t^2} 2tdt[/Tex]

[Tex]= 2\int_{\sqrt{2}}^{2}\left ( t^2-2 \right )t^2dt[/Tex]

[Tex]= 2\int_{\sqrt{2}}^{2}\left ( t^4-2t^2 \right )dt[/Tex]

[Tex]= 2\left [ \frac{t^5}{5}-\frac{2t^3}{3} \right ]_{\sqrt{2}}^{2}[/Tex]

[Tex]= 2\left [ \frac{32}{5}-\frac{16}{3}-\frac{4\sqrt{2}}{5}+\frac{4\sqrt{2}}{5} \right ][/Tex]

[Tex]= 2\left [ \frac{96-80-12\sqrt{2}+20\sqrt{2}}{15} \right ][/Tex]

[Tex]= 2\left [ \frac{16+8\sqrt{2}}{15} \right ][/Tex]

[Tex]= \frac{16\left ( 2+\sqrt{2}\right )}{15}[/Tex]

[Tex]= \frac{16\sqrt{2}\left ( \sqrt{2}+1\right )}{15}[/Tex]

Question 5

[Tex]\int_{0}^{\frac{\pi}{2}}\frac{sinx}{1+cos^{2}x}dx[/Tex]

Solution:

[Tex]\int_{0}^{\frac{\pi}{2}}\frac{sinx}{1+cos^{2}x}dx[/Tex]

Let [Tex]cosx=t \Rightarrow -sinx dx = dt[/Tex]

When [Tex]x=0, t=1 [/Tex]and when [Tex]x=\frac{\pi}{2}, t=0[/Tex]

[Tex]\Rightarrow \int_{0}^{\frac{\pi}{2}}\frac{sinx}{1+cos^{2}x}dx = -\int_{1}^{0}\frac{dt}{1+t^2}[/Tex]

[Tex]=-\left [ tan^{-1}t \right ]_{1}^{0}[/Tex]

[Tex]=-\left [ tan^{-1}0 – tan^{-1}1 \right ][/Tex]

[Tex]=-\left [ -\frac{\pi}{4} \right ][/Tex]

[Tex]=\frac{\pi}{4}[/Tex]

Question 6

[Tex]\int_{0}^{2} \frac{dx}{x+4-x^{2}}[/Tex]

Solution:

[Tex]\int_{0}^{2} \frac{dx}{x+4-x^{2}}=\int_{0}^{2} \frac{dx}{-\left (x^{2}-x-4 \right )}[/Tex]

[Tex]=\int_{0}^{2} \frac{dx}{-\left (x^{2}-x+\frac{1}{4}-\frac{1}{4}-4 \right )}[/Tex]

[Tex]=\int_{0}^{2} \frac{dx}{-\left [ \left ( x-\frac{1}{2} \right )^{2} -\frac{17}{2} \right ]}[/Tex]

[Tex]=\int_{0}^{2} \frac{dx}{ \left ( \frac{\sqrt{17}}{2} \right )^{2} – \left ( x-\frac{1}{2} \right )^{2} }[/Tex]

Let [Tex]x-\frac{1}{2}=t\Rightarrow dx=dt[/Tex]

When [Tex]x=0, t=-\frac{1}{2}[/Tex] and when [Tex]x=2[/Tex], [Tex]t=\frac{3}{2}[/Tex]

[Tex]\therefore =\int_{0}^{2} \frac{dx}{ \left ( \frac{\sqrt{17}}{2} \right )^{2} – \left ( x-\frac{1}{2} \right )^{2} } = \int_{0}^{2} \frac{dt}{ \left ( \frac{\sqrt{17}}{2} \right )^{2} – t^{2} }[/Tex]

[Tex]=\left [ \frac{1}{2\left ( \frac{\sqrt{17}}{2} \right )} log\frac{\frac{\sqrt{17}}{2}+t}{\frac{\sqrt{17}}{2}-t} \right ]_{-\frac{1}{2}}^{\frac{3}{2}}[/Tex]

[Tex]=\frac{1}{\sqrt{17}}\left [ log\frac{\frac{\sqrt{17}}{2}+\frac{3}{2}}{\frac{\sqrt{17}}{2}-\frac{3}{2}} – \frac{log\frac{\sqrt{17}}{2}-\frac{1}{2}}{log\frac{\sqrt{17}}{2}+\frac{1}{2}} \right ][/Tex]

[Tex]=\frac{1}{\sqrt{17}}\left [ log \frac{\sqrt{17}+3}{\sqrt{17}-3} – log \frac{\sqrt{17}-1}{\sqrt{17}+1} \right ][/Tex]

[Tex]=\frac{1}{\sqrt{17}} log \frac{\sqrt{17}+3}{\sqrt{17}-3} \times \frac{\sqrt{17}+1}{\sqrt{17}-1}[/Tex]

[Tex]=\frac{1}{\sqrt{17}} log \left [ \frac{17+3+4\sqrt{17}}{17+3-4\sqrt{17}} \right ][/Tex]

[Tex]=\frac{1}{\sqrt{17}} log \left [ \frac{20+4\sqrt{17}}{20-4\sqrt{17}} \right ][/Tex]

[Tex]=\frac{1}{\sqrt{17}} log \left ( \frac{5+\sqrt{17}}{5-\sqrt{17}} \right )[/Tex]

[Tex]=\frac{1}{\sqrt{17}} log \left [ \frac{(5+\sqrt{17})(5+\sqrt{17})}{25-17} \right ][/Tex]

[Tex]=\frac{1}{\sqrt{17}} log \left [ \frac{25+17+10\sqrt{17}}{8} \right ][/Tex]

[Tex]=\frac{1}{\sqrt{17}} log \left ( \frac{42+10\sqrt{17}}{8} \right )[/Tex]

[Tex]=\frac{1}{\sqrt{17}} log \left ( \frac{21+5\sqrt{17}}{4} \right )[/Tex]

Question 7

[Tex]\int_{-1}^{1} \frac{dx}{x^{2}+2x+5}[/Tex]

Solution:

[Tex]\int_{-1}^{1} \frac{dx}{x^{2}+2x+5}=\int_{-1}^{1} \frac{dx}{\left (x^{2}+2x +1 \right )+4}=\int_{-1}^{1} \frac{dx}{\left (x+1 \right )^2+(2)^2}[/Tex]

Let [Tex]x+1=t\Rightarrow dx=dt[/Tex]

When [Tex]x=-1, t=0[/Tex] and when [Tex]x=1, t=2[/Tex]

[Tex]\therefore \int_{-1}^{1} \frac{dx}{\left (x+1 \right )^2+(2)^2}=\int_{0}^{2} \frac{dt}{t^2 + 2^2}[/Tex]

[Tex]=\left [ \frac{1}{2}tan^{-1}\frac{1}{2} \right ]_{0}^{2}[/Tex]

[Tex]= \frac{1}{2}tan^{-1}1-\frac{1}{2}tan^{-1}0[/Tex]

[Tex]= \frac{1}{2}\left ( \frac{\pi}{4} \right )[/Tex]

[Tex]= \frac{\pi}{8}[/Tex]

Question 8.

[Tex]\int_{1}^{2}\left( \frac{1}{x}-\frac{1}{2x^{2}} \right)e^{2x}dx[/Tex]

Solution:

[Tex]\int_{1}^{2}\left( \frac{1}{x}-\frac{1}{2x^{2}} \right)e^{2x}dx[/Tex]

Let [Tex]2x=t\Rightarrow 2dx=dt[/Tex]

When [Tex]x=1, t=2[/Tex] and when [Tex]x=2, t=4[/Tex]

[Tex]\therefore \int_{1}^{2}\left( \frac{1}{x}-\frac{1}{2x^{2}} \right)e^{2x}dx = \frac{1}{2}\int_{2}^{4}\left( \frac{2}{t}-\frac{2}{t^{2}} \right)e^{t}dt = \int_{2}^{4}\left( \frac{1}{t}-\frac{1}{t^{2}} \right)e^{t}dt[/Tex]

Let [Tex]\frac{1}{t}=f(t)[/Tex]

Then, [Tex]f'(t)=\frac{1}{t^2}[/Tex]

[Tex]\Rightarrow = \int_{2}^{4}\left( \frac{1}{t}-\frac{1}{t^{2}} \right)e^{t}dt = \int_{2}^{4}e^t\left [ f(t)+f'(t) \right ]dt[/Tex]

[Tex]=\left [ e^t f(t) \right ]_{2}^{4}[/Tex]

[Tex]=\left [ e^t \cdot \frac{2}{t} \right ]_{2}^{4}[/Tex]

[Tex]=\left [ \frac{e^t}{t} \right ]_{2}^{4}[/Tex]

[Tex]=\frac{e^4}{4}-\frac{e^2}{2}[/Tex]

[Tex]=\frac{e^2(e^2-2)}{4}[/Tex]

Question 9.

Value of the integral [Tex]\int_{\frac{1}{3}}^{1}\frac{(x-x^{3})^{\frac{1}{3}}}{x^{4}}dx[/Tex] is:

• [A] 6
• [B] 0
• [C] 3
• [D] 4

Correct Answer is Option [A] 6

Solution:

Let I = [Tex]\int_{\frac{1}{3}}^{1}\frac{(x-x^{3})^{\frac{1}{3}}}{x^{4}}dx[/Tex]

= [Tex]\int_{\frac{1}{3}}^{1}[/Tex][x³(x^-2 -1)]^1/3/x⁴ dx

= [Tex]\int_{\frac{1}{3}}^{1}[/Tex] x(x^-2 – 1)^1/3/x⁴ dx

= [Tex]\int_{\frac{1}{3}}^{1}[/Tex] (x^-2 – 1)^1/3x^-3 dx…(i)

let, x^-2 – 1 = t

-2x^-3 = dt/dx

x^-3dx = -1/2dt

Changing limit of Integration

when, x = 1/3, t = x^-2 -1 = (1/3)^-2 – 1 = 8

when, x = 1, t = x^-2 -1 = (1)^-2 – 1 = 0

From (i)

I = -1/2 [Tex]\int_{8}^{0}[/Tex]t^1/3.dt

I = -1/2(t^4/3/{4/3})⁰₈

I = -1/2.3/4{0 – 16}

I = 6

Question 10.

If [Tex]f(x)=\int_{0}^{x}t sint dt[/Tex] , then f'(x) is

• [A] cosx + x sinx
• [B] x sinx
• [C] x cosx
• [D] sinx + x cosz

Correct Answer is Option [B] x sinx

Solution:

f(x) = [Tex]\int_{0}^{x}t sint dt[/Tex]…(i)

let,

• u = t
• v = cos t

du = dt

dv = -sint.dt

using, ∫udv = uv − ∫vdu

from eq(i)

= [-tcos(t)]^x₀ – [Tex]\int_{0}^{x}(-cost) dt[/Tex]

= [-x.cosx – (-0.cos0)} – [Tex]\int_{0}^{x}(-cost) dt[/Tex]

= -x.cosx + [Tex]\int_{0}^{x}cos(t). dt[/Tex]

= -x.cos(x) + [sin (t)]^x₀

= -x.cos(x) + sin(x) – sin (0)

Similar Reads:

• Class 12 NCERT Solutions- Mathematics Part ii – Chapter 7– Integrals Exercise 7.1
• Class 12 NCERT Solutions- Mathematics Part ii – Chapter 7– Integrals Exercise 7.3
• Class 12 NCERT Solutions- Mathematics Part ii – Chapter 7– Integrals Exercise 7.4
• Class 12 NCERT Solutions- Mathematics Part ii – Chapter 7– Integrals Exercise 7.5
• Class 12 NCERT Solutions- Mathematics Part ii – Chapter 7– Integrals Exercise 7.6
• Class 12 NCERT Solutions – Mathematics Part ii – Chapter 8 – Application of Integrals Miscellaneous Exercise