Mathematics

# Evaluate $\int_{0}^{1} x \tan^{-1} x\ dx$.

##### SOLUTION
$\displaystyle \int f(x) g(x) dx= f(x) \displaystyle \int g(x) dx= \displaystyle \int (f(x) \displaystyle \int g(x)dx)dx$
$\displaystyle \int_0^1 x\tan^{-1}x\ dx=\displaystyle \int_0^1 (\tan^{-1}x)x\ dx$
$=\tan^{-1}x \displaystyle \int_0^1x\ dx-\displaystyle \int_0^1\left (\dfrac {d(\tan^{-1}x)}{dx} \displaystyle \int x. dx\right) dx$
$=\left [\tan^{-1}x. \dfrac {x^2}{2}\right]_0^1 -\displaystyle \int_0^1 \dfrac {1}{1+x^2}dx$
$=\left [\dfrac {x^2}{2}\tan^{-1}x\right]-\dfrac {1}{2} \displaystyle \int_0^1 \dfrac {x^2 +1-1}{x^2 +1}dx$
$=\left [\dfrac {x^2}{2}\tan^{-1}x\right]_0^1 -\dfrac {1}{2} \left [\displaystyle \int_0^1 \dfrac {x^2 +1}{x^2 +1}dx -\displaystyle \int_0^1 \dfrac {dx}{x^2 +1}\right]$
$=\left [\dfrac {x^2}{2}\tan^{-1}x\right]-\dfrac {1}{2} \left [\displaystyle \int_0^1 dx-\displaystyle \int_0^1 \dfrac {dx}{x^2 +1}\right]$
$\left [Using\ \displaystyle \int \dfrac {dx}{a^2 +x^2}=\dfrac {1}{a} \tan^{-1} \dfrac {x}{a}+C\right]$
$=\left [\dfrac {x^2}{2}\tan^{-1}x-\dfrac {1}{2}x +\dfrac {1}{2}+\dfrac {1}{1}\tan^{-1}\dfrac {x}{1}\right]_0^1 C$
$=\left [\dfrac {x^2}{2}\tan^{-1}x -\dfrac {x}{2}+\dfrac {1}{2}\tan^{-1}x\right]_0^1$
$=\left[ \dfrac { { \left( 1 \right) }^{ 2 } }{ 2 } \tan ^{ 1 }{ \left( 1 \right) } -\dfrac { 1 }{ 2 } +\dfrac { 1 }{ 2 } \tan ^{ -1 }{ 1 } \right] -\left [\dfrac {0^2}{2}\tan^{-1}0-\dfrac {0}{2}+\dfrac {1}{2}\tan^{-1} (0)\right]$
$=\left [\dfrac {\pi}{8}-\dfrac {1}{2}+\dfrac {\pi}{8}\right]-[0-0+0]$
$= \pi /4 -\dfrac {1}{2}$

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 111

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