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Integrate the function $\sin^{-1}\bigg(\large\frac{2x}{1+x^2}\bigg)$

$\begin{array}{1 1} 2x \tan^{-1}x- \log |1+x^2|+c \\ 2x \cos^{-1}x- \log |1+x^2|+c \\2x \cos^{-1}x+ \log |1+x^2|+c \\ 2x \cos^{-1}x- \log |1-x^2|+c \end{array} $

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  • Let u and v be two functions,such that the integral function is of the form $\int vdu,$ then it can be solved by the method of integration by parts.
  • $\int udv=uv-\int vdu.$
  • $\frac{d}{dx}(\tan x)=\sec^2x$
  • $ \int \sec^2 x=\tan x+c.$
Step 1:
Let $I=\int\sin^{-1}\bigg(\large\frac{2x}{1+x^2}\bigg)$$dx$
Substitute $x=\tan\theta$ we get,
$I=\int \sin^{-1}\bigg(\large\frac{2\tan\theta}{1+\tan^2\theta}\bigg)$$dx$
We know that $\sin 2\theta=\large\frac{2\tan\theta}{1+\tan^2\theta}$
$\int \sin^{-1}(\sin 2\theta)dx$
$\Rightarrow \int\large\frac{1}{ \sin}\times$$ \sin2\theta dx$
$\Rightarrow 2\int \theta dx$
Step 2:
We have $x=\tan\theta$
By substituting the value of $\theta$ we get,
$2\int \tan^{-1}xdx=2\tan^{-1}x\int dx-2\int\large\frac{d}{dx}$$\tan^{-1}x\int dx]dx$
By using integration by parts,
$I=2\tan^{-1}x.x-2\int \bigg[\large\frac{1}{1+x^2}$$x\bigg]dx$
Step 3:
Put $1+x^2=t$
By differentiating with respect to $x$
$\Rightarrow 2x=\large\frac{dt}{dx}$
Step 4:
$I=2x\tan^{-1}x-2\int \bigg[\large\frac{x}{t}.\large\frac{dt}{2x}\bigg]$
$\;\;=2x\tan^{-1}x-\int \large\frac{1}{t}$$dt$
$\;\;=2x\tan^{-1}x-\log \mid t\mid+c$
$I=2x\tan ^{-1}x-\log\mid 1+x^2\mid+c$
answered Sep 12, 2013 by sreemathi.v
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