Mathematics

# Evaluate$\int \sin^{-1} \dfrac{2x}{1+x^2} dx$

##### SOLUTION
$I=\int \sin^{-1}\dfrac{2x}{1+x^{2}}dx$

Put $x=\tan t$

Then, $sin^{-1} \dfrac{2x}{1+x^{2}}=sin^{-1} \dfrac{2\tan t}{1+tan^{2}t}$

$=sin^{-1} sin2t$

$=2t$

Thus, our function becomes

$I=2\int tan^{-1}xdx$

$I=2tan^{-1}x \int 1. dx -2\int (\dfrac{d(tan^{-1}x)}{dx} \int 1.dx)dx$

$I=2xtan^{-1}x-2\int \dfrac{1}{1+x^{2}}xdx$

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

Let $1+x^{2}=t$

$dx=\dfrac{dt}{2x}$

Therefore,

$I_{1}=\int \dfrac{x}{t}. \dfrac{dt}{2x}$

$I_{1}=\dfrac{1}{2}\int \dfrac{dt}{t}$

$I_{1}=\dfrac{1}{2}\log t$

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

Hence,

$I=2xtan^{-1}x-\log (1+x^{2})+C$

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 86

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