Mathematics

Evaluate: $$\displaystyle\int {{sin^{ - 1}}} \left( {{{2x} \over {1 + {x^2}}}} \right)dx$$


SOLUTION
$$I=\displaystyle \int {\sin^{-1}\left(\dfrac{2x}{1+x^2}\right)}$$

Simplifying the given function 

$$\sin^{-1}\left(\dfrac{2x}{1+x^2}\right)$$

Put $$x=\tan t$$

$$\therefore t=\tan^{-1}(x)$$

$$\therefore \sin^{-1}\left(\dfrac{2x}{1+x^2}\right)=\sin^{-1}\left(\dfrac{2\tan t}{1+\tan^2t}\right)$$

                                  $$=\sin^{-1}(\sin 2t)$$

                                  $$=2t$$

                                  $$2\tan^{-1}x$$

$$\therefore I=\displaystyle \int{\sin^{-1}\left(\dfrac{2x}{1+x^2}\right)}=2\displaystyle \int{\tan^{-1}x dx}$$

                                              $$=2\tan^{-1}x.x-2\displaystyle \int{\dfrac{x}{1+x^2}dx}$$

                                              $$=2x\tan^{-1}x-\log|1+x^2|+c$$
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Subjective Medium Published on 17th 09, 2020
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