CBSE Class 12 Maths Chapter 7 Integrals – NCERT Solutions 2025–26

Exercise 7.8

1.Integrate the given integral $\mathbf{\int\limits_{ - 1}^1 {(x + 1)dx} }$.

Ans: To simplify the question let us suppose, $I = \int\limits_{ - 1}^1 {(x + 1)dx} $

$\int {(x + 1)dx = \dfrac{{{x^2}}}{2}}  + x$

Again, let us suppose the function is $F(x) = \dfrac{{{x^2}}}{2} + x$

By second fundamental theorem of calculus, we obtain

$  I = F(1) - F( - 1) $

$= \left( {\dfrac{1}{2} + 1} \right) - \left( {\dfrac{1}{2} - 1} \right) $

$   = \dfrac{1}{2} + 1 - \dfrac{1}{2} + 1 $

$   = 2 $ 

2.Integrate the given integral $\mathbf{\int\limits_2^3 {\dfrac{1}{x}dx} }$. 

Ans: To simplify the question let us suppose, $I = \int\limits_2^3 {\dfrac{1}{x}dx} $

Thus, $\int {\dfrac{1}{x}dx = \log \left| x \right|} $

Again, let us suppose the function is$F(x) = \log \left| x \right|$

By second fundamental theorem of calculus, we obtain

$  I = F(3) - F(2) $

$ = \log \left| 3 \right| - \log \left| 2 \right| $

$   = \log \dfrac{3}{2} $ 

3.Integrate the given integral $\mathbf{\int\limits_1^2 {\left( {4{x^3} - 5{x^2} + 6x + 9} \right)dx} }$

Ans: To simplify the question let us suppose,

$I = \int\limits_1^2 {\left( {4{x^3} - 5{x^2} + 6x + 9} \right)dx}  $

$\int\limits_1^2 {\left( {4{x^3} - 5{x^2} + 6x + 9} \right)dx}  = 4\left( {\dfrac{{{x^4}}}{4}} \right) - 5\left( {\dfrac{{{x^3}}}{3}} \right) + 6\left( {\dfrac{{{x^2}}}{2}} \right) + 9(x) $ 

Again, let us suppose the function is 

$ \Rightarrow {x^4} - \dfrac{{5{x^3}}}{3} + 3{x^2} + 9x = F(x)$

By second fundamental theorem of calculus, we obtain

$I = F(2) - F(1)$

$ I = \left\{ {{2^4} - \dfrac{{5{{(2)}^3}}}{3} + 3{{(2)}^2} + 9(2)} \right\} - \left\{ {{{(1)}^4} - \dfrac{{5{{(1)}^3}}}{3} + 3{{(1)}^2} + 9(1)} \right\} $

$ = \left( {16 - \dfrac{{40}}{3} + 12 + 18} \right) - \left( {1 - \dfrac{5}{3} + 3 + 9} \right) $

$ = 16 - \dfrac{{40}}{3} + 12 + 18 - 1 + \dfrac{5}{3} - 3 - 9 $

$  = 33 - \dfrac{{35}}{3} $

$ = \dfrac{{99 - 35}}{3} $

$ = \dfrac{{64}}{3} $

4.Integrate the given integral $\mathbf{\int\limits_0^{\dfrac{\pi }{4}} {\left( {\sin 2xdx} \right)}} $

Ans: To simplify the question let us suppose,

$I = \int\limits_0^{\dfrac{\pi }{4}} {\left( {\sin 2xdx} \right)}  $

  $\int {\sin 2xdx = \left( {\dfrac{{ - \cos 2x}}{2}} \right)}  $ 

Again, let us suppose the function is$\left( {\dfrac{{ - \cos 2x}}{2}} \right) = F(x)$

By second fundamental theorem of calculus, we obtain

$  I = F\left( {\dfrac{\pi }{4}} \right) - F(0) $

$  =  - \dfrac{1}{2}\left[ {\cos 2\left( {\dfrac{\pi }{4}} \right) - \cos 0} \right] $

$  =  - \dfrac{1}{2}\left[ {\cos 2\left( {\dfrac{\pi }{4}} \right) - \cos 0} \right] $

$   =  - \dfrac{1}{2}\left[ {\cos 2\left( {\dfrac{\pi }{4}} \right) - \cos 0} \right] $

$  =  - \dfrac{1}{2}\left[ {0 - 1} \right] $

$   = \dfrac{1}{2} $ 

5.Integrate the given integral $\mathbf{\int\limits_0^{\dfrac{\pi }{2}} {\left( {\cos 2xdx} \right)} }$

Ans: To simplify the question let us suppose,

$  I = \int\limits_0^{\dfrac{\pi }{2}} {\left( {\cos 2xdx} \right)}  $

$  \int {\cos 2xdx = \left( {\dfrac{{\sin 2x}}{2}} \right)}  $ 

Again, let us suppose the function is$\left( {\dfrac{{\sin 2x}}{2}} \right) = F(x)$

By second fundamental theorem of calculus, we obtain

$  I = F\left( {\dfrac{\pi }{2}} \right) - F(0) $

$   = \dfrac{1}{2}\left[ {\sin 2\left( {\dfrac{\pi }{2}} \right) - \sin 0} \right] $

$   = \dfrac{1}{2}\left[ {\sin \pi  - \sin 0} \right] $

$   = \dfrac{1}{2}\left[ {0 - 0} \right] $

$   = 0 $

6.Integrate the given integral $\mathbf{\int\limits_4^5 {{e^x}dx} }$

Ans: To simplify the question let us suppose,

$  I = \int\limits_4^5 {{e^x}dx}  $

$  \int\limits_4^5 {{e^x}dx}  = {e^x} = F(x) $ 

By second fundamental theorem of calculus, we obtain

$I = F(5) - F(4) $

$ = {e^5} - {e^4} $

 $  = {e^4}\left( {e - 1} \right) $ 

7.Integrate the given integral $\mathbf{\int\limits_0^{\dfrac{\pi }{4}} {\tan xdx} }$

Ans: To simplify the question let us suppose,

$I = \int\limits_0^{\dfrac{\pi }{4}} {\tan xdx}  $

 $ \int {\tan xdx =  - \log \left| {\cos x} \right|}  $ 

Again, let us suppose the function is$ - \log \left| {\cos x} \right| = F(x)$

By second fundamental theorem of calculus, we obtain

$ I = F\left( {\dfrac{\pi }{4}} \right) - F(0) $

$  =  - \log \left| {\cos \dfrac{\pi }{4}} \right| + \log \left| {\cos 0} \right| $

$   =  - \log \left| {\dfrac{1}{{\sqrt 2 }}} \right| + \log \left| 1 \right| $

$   =  - \log {\left( 2 \right)^{ - \dfrac{1}{2}}} $

$   = \dfrac{1}{2}\log 2 $ 

8.Integrate the given integral $\mathbf{\int\limits_{\dfrac{\pi }{6}}^{\dfrac{\pi }{4}} {\cos ecxdx} }$

Ans: To simplify the question let us suppose,

$ I = \int\limits_{\dfrac{\pi }{6}}^{\dfrac{\pi }{4}} {\cos ecxdx}  $

$  \int {\cos ecxdx = \log \left| {\cos ecx\dfrac{\pi }{6} - \cot \dfrac{\pi }{6}} \right|}  $ 

Again, let us suppose the function is$F(x) = \log \left| {\cos x} \right|$

By second fundamental theorem of calculus, we obtain

$I = F\left( {\dfrac{\pi }{4}} \right) - F\left( {\dfrac{\pi }{6}} \right) $

$   = \log \left| {\cos ec\dfrac{\pi }{4} - \cot \dfrac{\pi }{4}} \right| - \log \left| {\cos ec\dfrac{\pi }{6} - \cot \dfrac{\pi }{6}} \right| $

$   = \log \left| {\sqrt 2  - 1} \right| - \log \left| {2 - \sqrt 3 } \right| $

$   = \log \left( {\dfrac{{\sqrt 2  - 1}}{{2 - \sqrt 3 }}} \right) $ 

9.Integrate the given integral $\mathbf{\int\limits_0^1 {\dfrac{{dx}}{{\sqrt {1 - {x^2}} }}} }$

Ans: To simplify the question let us suppose,

$I = \int\limits_0^1 {\dfrac{{dx}}{{\sqrt {1 - {x^2}} }}} $

Again, let us suppose the function is

$ I = \int {\dfrac{{dx}}{{\sqrt {1 - {x^2}} }} = F(x)}  $

$ {\sin ^{ - 1}}x = F(x) $ 

By second fundamental theorem of calculus, we obtain

$ I = F(1) - F(0) $

$  = {\sin ^{ - 1}}(1) - {\sin ^{ - 1}}(0) $

$ = \dfrac{\pi }{2} - 0 $

$   = \dfrac{\pi }{2} $ 

10.Integrate the given integral $\mathbf{\int\limits_0^1 {\dfrac{{dx}}{{\sqrt {1 + {x^2}} }}}} $

Ans: To simplify the question let us suppose,

$  I = \int\limits_0^1 {\dfrac{{dx}}{{1 + {x^2}}}}  $

$  \int {\dfrac{{dx}}{{1 + {x^2}}} = {{\tan }^{ - 1}}}  $ 

Again, let us suppose the function is

${\tan ^{ - 1}}x = F(x)$

By second fundamental theorem of calculus, we obtain

$I = F(1) - F(0) $

$= {\tan ^{ - 1}}(1) - {\tan ^{ - 1}}(0) $

$ = \dfrac{\pi }{4} $ 

11.Integrate the given integral $\mathbf{\int\limits_2^3 {\dfrac{{dx}}{{{x^2} - 1}}} }$

Ans: To simplify the question let us suppose,

$  I = \int\limits_2^3 {\dfrac{{dx}}{{{x^2} - 1}}}  $

$ \int {\dfrac{{dx}}{{{x^2} - 1}} = \dfrac{1}{2}\log \left| {\dfrac{{x - 1}}{{x + 1}}} \right|}  $ 

Again, let us suppose the function is

$\dfrac{1}{2}\log \left| {\dfrac{{x - 1}}{{x + 1}}} \right| = F(x)$

By second fundamental theorem of calculus, we obtain

$  I = F(3) - F(2) $

$   = \dfrac{1}{2}\left[ {\log \left| {\dfrac{{3 - 1}}{{3 + 1}}} \right| - \log \left| {\dfrac{{2 - 1}}{{2 + 1}}} \right|} \right] $

$   = \dfrac{1}{2}\left[ {\log \left| {\dfrac{2}{4}} \right| - \log \left| {\dfrac{1}{3}} \right|} \right] $

$   = \dfrac{1}{2}\left[ {\log \dfrac{1}{2} - \log \dfrac{1}{3}} \right] $

$  = \dfrac{1}{2}\left[ {\log \dfrac{3}{2}} \right] $ 

12.Integrate the given integral $\mathbf{\int\limits_0^{\dfrac{\pi }{2}} {{{\cos }^2}xdx} }$

Ans: To simplify the question let us suppose,

$\int {{{\cos }^2}xdx = \int {\left( {\dfrac{{1 + \cos 2x}}{2}} \right)dx} }  $

$ \dfrac{x}{2} + \dfrac{{\sin 2x}}{4} = \dfrac{1}{2}\left( {x + \dfrac{{\sin 2x}}{2}} \right) $ 

Again, let us suppose the function is

$\dfrac{1}{2}\left( {x + \dfrac{{\sin 2x}}{2}} \right) = F(x)$

By second fundamental theorem of calculus, we obtain

$  I = F\left( {\dfrac{\pi }{2}} \right) - F(0) $

$ = \dfrac{1}{2}\left[ {\left( {\dfrac{\pi }{2} + \dfrac{{\sin \pi }}{2}} \right) - \left( {0 + \dfrac{{\sin 0}}{2}} \right)} \right] $

$   = \dfrac{1}{2}\left[ {\dfrac{\pi }{2} + 0 - 0 + 0} \right] $

$   = \dfrac{\pi }{4} $ 

13.Integrate the given integral $\mathbf{\int\limits_2^3 {\dfrac{{xdx}}{{{x^2} + 1}}}} $

Ans: To simplify the question let us suppose,

$ I = \int\limits_2^3 {\dfrac{{xdx}}{{{x^2} + 1}}}  $

$  \int {\dfrac{{xdx}}{{{x^2} + 1}}}  $

$   = \dfrac{1}{2}\int {\dfrac{{2x}}{{{x^2} + 1}}dx}  $

$   = \dfrac{1}{2}\log \left( {1 + {x^2}} \right) $ 

Again, let us suppose the function is

$\dfrac{1}{2}\log \left( {1 + {x^2}} \right) = F(x)$

By second fundamental theorem of calculus, we obtain

$  I = F(3) - F(2) $

$ = \dfrac{1}{2}\left[ {\log \left( {1 + {{\left( 3 \right)}^2}} \right) - \log \left( {1 + {{(2)}^2}} \right)} \right] $

$   = \dfrac{1}{2}\left[ {\log 10 - \log 5} \right] $

$ = \dfrac{1}{2}\left[ {\log \dfrac{{10}}{5}} \right] $

$= \dfrac{1}{2}\left[ {\log 2} \right] $ 

14.Integrate the given integral $\mathbf{\int\limits_0^1 {\dfrac{{2x + 3}}{{5{x^2} + 1}}dx} }$

Ans: To simplify the question let us suppose,

$  I = \int\limits_0^1 {\dfrac{{2x + 3}}{{5{x^2} + 1}}dx}  $

$  \int {\dfrac{{2x + 3}}{{5{x^2} + 1}}dx}  $

$ = \dfrac{1}{5}\int {\dfrac{{5\left( {2x + 3} \right)}}{{5{x^2} + 1}}dx}  $

$   = \dfrac{1}{5}\int {\dfrac{{\left( {10x + 15} \right)}}{{5{x^2} + 1}}dx}  $

$  = \dfrac{1}{5}\int {\dfrac{{10x}}{{5{x^2} + 1}}dx}  + 3\int {\dfrac{1}{{5{x^2} + 1}}dx}  $

$= \dfrac{1}{5}\int {\dfrac{{10x}}{{5{x^2} + 1}}dx}  + 3\int {\dfrac{1}{{5\left( {{x^2} + \dfrac{1}{5}} \right)}}dx}  $

$   = \dfrac{1}{5}\log \left( {5{x^2} + 1} \right) + \dfrac{3}{5}\dfrac{1}{{\dfrac{1}{{\sqrt 5 }}}}{\tan ^{ - 1}}\dfrac{x}{{\dfrac{1}{{\sqrt 5 }}}} $

$   = \dfrac{1}{5}\log \left( {5{x^2} + 1} \right) + \dfrac{3}{{\sqrt 5 }}{\tan ^{ - 1}}\left( {\sqrt 5 } \right)x $

$ = F(x) $ 

By second fundamental theorem of calculus, we obtain

$  I = F(3) - F(2) $

$   = \dfrac{1}{5}\left[ {\log \left( {1 + 5} \right) - \dfrac{3}{{\sqrt 5 }}\log \left( {\sqrt 5 } \right)} \right] - \left[ {\dfrac{1}{5}\log (1) + \dfrac{3}{{\sqrt 5 }}{{\tan }^{ - 1}}(0)} \right] $

$   = \dfrac{1}{5}\left[ {\log 6} \right] + \dfrac{3}{{\sqrt 5 }}{\tan ^{ - 1}}\sqrt 5 $ 

15.Integrate the given integral $\mathbf{\int\limits_0^1 {x{e^{{x^2}}}dx} }$

Ans: To simplify the question let us suppose,

$= \int\limits_0^1 {x{e^{{x^2}}}dx}$

$  {x^2} = t $

$ \Rightarrow 2xdx = dt $ 

As $x \to 0,t \to 0$ ,

Again $x \to 1,t \to 1,$, 

$\therefore I = \dfrac{1}{2}\int\limits_0^1 {{e^t}dt}  $

$  \dfrac{1}{2}\int\limits_0^1 {{e^t}dt}  = \dfrac{1}{2}{e^t}dt $

$  \dfrac{1}{2}{e^t}dt = F(t) $ 

By second fundamental theorem of calculus, we obtain

$  I = F(1) - F(0) $

$   = \dfrac{1}{2}e - \dfrac{1}{2}{e^0} $

$   = \dfrac{1}{2}\left( {e - 1} \right) $ 

16.Integrate the given integral $\mathbf{\int\limits_0^1 {\dfrac{{5{x^2}}}{{{x^2} + 4x + 3}}dx}} $

Ans: To simplify the question let us suppose,

$I = \int\limits_0^1 {\dfrac{{5{x^2}}}{{{x^2} + 4x + 3}}dx} $

Dividing $5{x^2}$by ${x^2} + 4x + 3$, we obtain

$ I = \int\limits_1^2 {\left\{ {5 - \dfrac{{20x + 15}}{{{x^2} + 4x + 3}}} \right\}} dx $

$ = \int\limits_1^2 {5dx - \int\limits_1^2 {\dfrac{{20x + 15}}{{{x^2} + 4x + 3}}} } dx $

$ = \left[ {5x} \right]_1^2 - \int\limits_1^2 {\dfrac{{20x + 15}}{{{x^2} + 4x + 3}}dx}  $

$  I = 5 - {I_1} $ 

 Consider,

Let,

$20x + 15 = A\dfrac{d}{{dx}}\left( {{x^2} + 4x + 3} \right) + B $

$= 2Ax + (4A + B) $ 

Equating the coefficients of $x$ and constant term, we obtain

$A = 10$ and $B =  - 25$

Let,${x^2} + 4x + 3 = t $

$  \Rightarrow \left( {2x + 4} \right)dx = dt $

$   \Rightarrow {I_1} = 10\int {\dfrac{{dt}}{t} - 25\int {\dfrac{{dx}}{{{{\left( {x + 2} \right)}^2} - {1^2}}}} }  $

$  = 10\log t - 25\left[ {\dfrac{1}{2}\log \left( {\dfrac{{x + 1}}{{x + 3}}} \right)} \right]_1^2 $

$  = \left[ {10\log 15 - 10\log 18} \right] - 25\left[ {\dfrac{1}{2}\log \dfrac{3}{5} - \dfrac{1}{2}\log \dfrac{2}{4}} \right] $

$  = \left[ {10\log 5 + 10\log 3 - 10\log 4 - 10\log 2} \right] - \dfrac{{25}}{2}\left[ {\log 3 - \log 5 - \log 2 + \log 4} \right] $

$  = \left[ {10 + \dfrac{{25}}{2}} \right]\log 5 + \left[ { - 10 - \dfrac{{25}}{2}} \right]\log 4 + \left[ {10 + \dfrac{{25}}{2}} \right]\log 3 + \left[ { - 10 + \dfrac{{25}}{2}} \right]\log 2 $

$   = \dfrac{{45}}{2}\log 5 - \dfrac{{45}}{2}\log 4 - \dfrac{5}{2}\log 3 + \dfrac{5}{2}\log 2 $

$   = \dfrac{{45}}{2}\log \dfrac{5}{4} - \dfrac{5}{2}\log \dfrac{3}{2} $ 

Substituting the value of ${I_1}$in (1), we obtain

$ I = 5 - \left[ {\dfrac{{45}}{2}\log \dfrac{5}{2} - \dfrac{5}{2}\log \dfrac{3}{2}} \right] $

$ = 5 - \dfrac{5}{2}\left[ {9\log \dfrac{5}{2} - \log \dfrac{3}{2}} \right] $ 

17.Integrate the given integral $\mathbf{\int\limits_0^{\dfrac{\pi }{4}} {\left( {2{{\sec }^2}x + {x^3} + 2} \right)dx} }$

Ans: To simplify the question let us suppose,

$ I = \int\limits_0^{\dfrac{\pi }{4}} {\left( {2{{\sec }^2}x + {x^3} + 2} \right)dx}  $

 $ \int {\left( {2{{\sec }^2}x + {x^3} + 2} \right)dx = 2\tan x + \dfrac{{{x^4}}}{4} + 2x}  $

$  2\tan x + \dfrac{{{x^4}}}{4} + 2x = F(x) $ 

By second fundamental theorem of calculus, we obtain

$  I = F\left( {\dfrac{\pi }{4}} \right) + F\left( 0 \right) $

$   = \left\{ {\left( {2\tan \dfrac{\pi }{4} + \dfrac{1}{4}{{\left( {\dfrac{\pi }{4}} \right)}^2} + 2\left( {\dfrac{\pi }{4}} \right)} \right) - \left( {2\tan 0 + 0 + 0} \right)} \right\} $

  $ = 2\tan \dfrac{\pi }{4} + {\dfrac{\pi }{{{4^5}}}^4} + \dfrac{\pi }{2} $

   $= 2 + \dfrac{\pi }{2} + \dfrac{{{\pi ^4}}}{{1024}} $ 

18.Integrate the given integral $\mathbf{\int\limits_0^\pi  {\left( {{{\sin }^2}\dfrac{x}{2} - {{\cos }^2}\dfrac{x}{2}} \right)dx} }$

Ans: To simplify the question let us suppose,

$ I = \int\limits_0^\pi  {\left( {{{\sin }^2}\dfrac{x}{2} - {{\cos }^2}\dfrac{x}{2}} \right)dx}  $

 $  =  - \int\limits_0^\pi  {\left( {{{\cos }^2}\dfrac{x}{2} - {{\sin }^2}\dfrac{x}{2}} \right)dx}  $

   $=  - \int\limits_0^\pi  {\cos xdx}  $

   $=  - \int\limits_0^\pi  {\cos xdx}  =  - \sin x $ 

By second fundamental theorem of calculus, we obtain

$  I = F\left( {\dfrac{\pi }{4}} \right) + F\left( 0 \right) $

  $ =  - \sin \pi  + \sin 0 $

$   = 0 $ 

19.Integrate the given integral $\mathbf{\int\limits_0^1 {\dfrac{{6x + 3}}{{{x^2} + 4}}dx} }$

Ans: To simplify the question let us suppose,

$ I = \int\limits_0^2 {\dfrac{{6x + 3}}{{{x^2} + 4}}dx}  $

$  \int {\dfrac{{6x + 3}}{{{x^2} + 4}}dx}  $

$   = 3\int {\dfrac{{2x + 1}}{{{x^2} + 4}}dx}  $

$   = 3\int {\dfrac{{2x + 1}}{{{x^2} + 4}}dx}  $

$   = 3\int {\dfrac{{2x}}{{{x^2} + 4}}dx}  + 3\int {\dfrac{1}{{{x^2} + 4}}dx}  $

$   = 3\log ({x^2} + 4) + \dfrac{3}{2}{\tan ^{ - 1}}\dfrac{x}{2} $

$ = F(x) $ 

By second fundamental theorem of calculus, we obtain

$ I = F(3) - F(2) $

$= 3\left[ {\log \left( {{2^2} + 4} \right) - \dfrac{3}{2}{{\tan }^{ - 1}}\left( {\dfrac{2}{2}} \right)} \right] - \left[ {3\log (0 + 4) + \dfrac{3}{2}{{\tan }^{ - 1}}(\dfrac{0}{2})} \right] $

$  = 3\left[ {\log 8} \right] + \dfrac{3}{2}{\tan ^{ - 1}}1 - 3\log 4 - \dfrac{3}{2}{\tan ^{ - 1}}0 $

$  = 3\log 8 + \dfrac{3}{2}{\tan ^{ - 1}}1 - 3\log 4 - \dfrac{3}{2}{\tan ^{ - 1}}0 $

$= 3\log 8 + \dfrac{3}{2}\left( {\dfrac{\pi }{4}} \right) - 3\log 4 - 0 $

$= 3\log \left( {\dfrac{8}{4}} \right) + \dfrac{{3\pi }}{8} $

$= 3\log 2 + \dfrac{{3\pi }}{8} $ 

20.Integrate the given integral $\int\limits_0^1 {\left( {x{e^x} - \sin \dfrac{{\pi x}}{4}} \right)dx} $

Ans: To simplify the question let us suppose,

$I = \int\limits_0^1 {\left( {x{e^x} - \sin \dfrac{{\pi x}}{4}} \right)dx}  $

$  \int\limits_0^1 {\left( {x{e^x} - \sin \dfrac{{\pi x}}{4}} \right)dx}  = x\int {{e^x}dx - \int {\left\{ {\left( {\dfrac{d}{{dx}}x} \right)\int {{e^x}dx} } \right\}dx + \left\{ {\dfrac{{ - \cos \dfrac{{\pi x}}{4}}}{{\dfrac{\pi }{4}}}} \right\}} }  $

$ = x{e^x} - x\int {{e^x}dx - \dfrac{4}{\pi }\cos \pi \dfrac{x}{4}}  $

$   = F(x) $ 

By second fundamental theorem of calculus, we obtain

 $ I = F(1) - F(0) $

 $= \left( {{e^1} - {e^1} - \dfrac{4}{\pi }\cos \pi \dfrac{1}{4}} \right) - \left( {0.{e^0} - {e^0} - \dfrac{4}{\pi }\cos 0} \right) $

 $  = e - e - \dfrac{4}{\pi }\left( {\dfrac{1}{{\sqrt 2 }}} \right) + 1 + \dfrac{4}{\pi } $

$= 1 + \dfrac{4}{\pi } - \dfrac{{2\sqrt 2 }}{\pi } $ 

21.Integrate the given integral $\int\limits_1^{\sqrt 3 } {\dfrac{{dx}}{{1 + {x^2}}}} $

1. \[\dfrac{\pi }{3}\]

2. $\dfrac{{2\pi }}{3}$

3. $\dfrac{\pi }{6}$

4. $\dfrac{\pi }{{12}}$

Ans: To simplify the question let us suppose,

$  \int {\dfrac{{dx}}{{1 + {x^2}}}}  = {\tan ^{ - 1}}x $

By second fundamental theorem of calculus, we obtain

$ \int\limits_1^{\sqrt 3 } {\dfrac{{dx}}{{1 + {x^2}}}}  = F(\sqrt 3 ) - F(1) $

$  = {\tan ^{ - 1}}\sqrt 3  - {\tan ^{^{ - 1}}}1 $

$= \dfrac{\pi }{3} - \dfrac{\pi }{4} $

$   = \dfrac{\pi }{{12}} $ 

Hence, the correct Answer is $\dfrac{\pi }{{12}}$

22.Integrate the given integral$\int {\dfrac{{dx}}{{4 + 9{x^2}}}}  = \int {\dfrac{{dx}}{{{{(2)}^2} + {{(3x)}^2}}}}$ 

1. $\dfrac{\pi }{6} $

2. $\dfrac{\pi }{{12}} $

3. $\dfrac{\pi }{{24}} $

4. $\dfrac{\pi }{4} $ 

equals

Ans: To simplify the question let us suppose,

 3x = t

$\Rightarrow 3dx = dt$ 

$\int {\dfrac{{dx}}{{{{(2)}^2} + {{(3x)}^2}}}}  = \dfrac{1}{3}\int {\dfrac{{dt}}{{{{(2)}^2} + {{(t)}^2}}}}  $

$ = \dfrac{1}{3}\left[ {\dfrac{1}{2}{{\tan }^{ - 1}}\dfrac{t}{2}} \right] $

$ = \dfrac{1}{6}{\tan ^{ - 1}}\left( {\dfrac{{3x}}{2}} \right) $

$= F(x) $ 

By second fundamental theorem of calculus,

$\int\limits_0^{\dfrac{2}{3}} {\dfrac{{dx}}{{4 + 9{x^2}}}}  = F\left( {\dfrac{2}{3}} \right) - F(0) $

$= \dfrac{1}{6}{\tan ^{ - 1}}\left( {\dfrac{3}{2} \times \dfrac{2}{3}} \right) - \dfrac{1}{6}{\tan ^{^{ - 1}}}0 $

 $  = \dfrac{1}{6}{\tan ^{ - 1}}1 - 0 $

 $  = \dfrac{1}{6} \times \dfrac{\pi }{4} $

 $  = \dfrac{\pi }{{24}} $ 

Hence, the correct answer is C.$\dfrac{\pi }{{24}}$

Conclusion

NCERT Solutions for Exercise 7.8 Maths Class 12 Chapter 7 - Integrals provide detailed explanations to all of the questions in the NCERT textbook. This exercise contains 22 questions with fully solved solutions. It includes a range of exercises that help students understand how integrals can be applied in real-world circumstances. With a focus on practical integration methods and their applications, Exercise 7.8 provides students with the knowledge they require to succeed in mathematics exams.

Class 12 Maths Chapter 7: Exercises Breakdown

CBSE Class 12 Maths Chapter 7 Other Study Materials

NCERT Solutions for Class 12 Maths | Chapter-wise List

Given below are the chapter-wise NCERT Class 12 Maths solutions PDF. Using these chapter-wise class 12th maths ncert solutions, you can get clear understanding of the concepts from all chapters.

Related Links for NCERT Class 12 Maths in Hindi

Explore these essential links for NCERT Class 12 Maths in Hindi, providing detailed solutions, explanations, and study resources to help students excel in their mathematics exams.

Important Related Links for NCERT Class 12 Maths