Chapter 7: Integrals

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] $ 

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} $ 

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] $ 

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 $ 

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) $ 

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] $ 

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}} $ 

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 $ 

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} $ 

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 } $