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.

Evaluate Integrals 1 to 8 Using Substitution

Question 1

\int_{0}^{1}\frac{x}{x^{2}+1}dx

Solution:

\int_{0}^{1}\frac{x}{x^{2}+1}

Let x^{2}+1=t \rightarrow 2x dx = dt

When x=0, t=1 and when x=1, t=2

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

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

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

=\frac{1}{2}log2

Question 2

\int_{0}^{\frac{\pi}{2}}\sqrt{sin\phi } cos^{5}\phi d\phi

Solution:

Let 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

Also, let sin \phi=t \Rightarrow cos\phi d\phi = dt

When \phi=0, t=0 and when \phi=\frac{\pi}{2}, t=1

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

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

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

=\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}

=\frac{2}{3}+\frac{2}{11}-\frac{4}{7}

=\frac{154+42-132}{231}

=\frac{64}{231}

Question 3

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

Solution:

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

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

When x=0, \theta=0 and when x=1, \theta=\frac{\pi}{4}

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

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

=\int_{0}^{\frac{\pi}{4}}2\theta \cdot sec^{2}\theta d\theta

2\int_{0}^{\frac{\pi}{4}}\theta \cdot sec^{2}\theta d\theta

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

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}}

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

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

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

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

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

=\frac{\pi}{2} - log2

Question 4

\int_{0}^{2}x\sqrt{x+2} (Put x+2 = t² )

Solution:

\int_{0}^{2}x\sqrt{x+2}dx

Let x+2 = t^2 and dx = 2tdt

When x=0, t=\sqrt{2} and when x=2, t=2

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

= 2\int_{\sqrt{2}}^{2}\left ( t^2-2 \right )t^2dt

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

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

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

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

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

= \frac{16\left ( 2+\sqrt{2}\right )}{15}

= \frac{16\sqrt{2}\left ( \sqrt{2}+1\right )}{15}

Question 5

\int_{0}^{\frac{\pi}{2}}\frac{sinx}{1+cos^{2}x}dx

Solution:

\int_{0}^{\frac{\pi}{2}}\frac{sinx}{1+cos^{2}x}dx

Let cosx=t \Rightarrow -sinx dx = dt

When x=0, t=1 and when x=\frac{\pi}{2}, t=0

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

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

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

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

=\frac{\pi}{4}

Question 6

\int_{0}^{2} \frac{dx}{x+4-x^{2}}

Solution:

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

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

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

=\int_{0}^{2} \frac{dx}{ \left ( \frac{\sqrt{17}}{2} \right )^{2} - \left ( x-\frac{1}{2} \right )^{2} }

Let x-\frac{1}{2}=t\Rightarrow dx=dt

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

\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} }

=\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}}

=\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 ]

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

=\frac{1}{\sqrt{17}} log \frac{\sqrt{17}+3}{\sqrt{17}-3} \times \frac{\sqrt{17}+1}{\sqrt{17}-1}

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

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

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

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

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

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

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

Question 7

\int_{-1}^{1} \frac{dx}{x^{2}+2x+5}

Solution:

\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}

Let x+1=t\Rightarrow dx=dt

When x=-1, t=0 and when x=1, t=2

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

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

= \frac{1}{2}tan^{-1}1-\frac{1}{2}tan^{-1}0

= \frac{1}{2}\left ( \frac{\pi}{4} \right )

= \frac{\pi}{8}

Question 8.

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

Solution:

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

Let 2x=t\Rightarrow 2dx=dt

When x=1, t=2 and when x=2, t=4

\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

Let \frac{1}{t}=f(t)

Then, f'(t)=\frac{1}{t^2}

\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

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

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

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

=\frac{e^4}{4}-\frac{e^2}{2}

=\frac{e^2(e^2-2)}{4}

Choose Correct Answer in Question 9 and 10

Question 9.

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

• [A] 6
• [B] 0
• [C] 3
• [D] 4

Correct Answer is Option [A] 6

Solution:

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

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

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

= \int_{\frac{1}{3}}^{1} (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 \int_{8}^{0}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 f(x)=\int_{0}^{x}t sint dt , 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) = \int_{0}^{x}t sint dt...(i)

let,

• u = t
• v = cos t

du = dt

dv = -sint.dt

using, ∫udv = uv − ∫vdu

from eq(i)

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

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

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

= -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