Question

Evaluate

$\int {\sin }^{-1}\frac{2x}{1+x^2}dx$

Answer 1

$I=\int {\sin }^{-1}\frac{2x}{1+x^2}dx$

Put $x=\tan t$

Then, $si{n}^{-1}\frac{2x}{1+x^2}=si{n}^{-1}\frac{2\tan t}{1+ta{n}^{2}t}$

$=si{n}^{-1}sin2t$

$=2t$

Thus, our function becomes

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

$I=2ta{n}^{-1}x\int 1.dx-2\int (\frac{d\left(ta{n}^{-1}x\right)}{dx}\int 1.dx)dx$

$I=2xta{n}^{-1}x-2\int \frac{1}{1+x^2}xdx$

Let $I_1=\int \frac{x}{1+x^2}dx$

Let $1+x^2=t$

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

Therefore,

$I_1=\int \frac{x}{t}.\frac{dt}{2x}$

$I_1=\frac{1}{2}\int \frac{dt}{t}$

$I_1=\frac{1}{2}\log t$

$I_1=\frac{1}{2}\log (1+x^2)$

Hence,

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