Mathematics

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


ANSWER

$$\displaystyle \log \left ( 1+\sin ^{2}x \right ).$$


SOLUTION
Let $$ \displaystyle I=\int  \frac { \sin  2x }{ 1+\sin ^{ 2 } x } dx$$
Put $$ \displaystyle 1+\sin  x=t\Rightarrow 2\sin  x\cos  xdx=dt$$
Therefore 
$$ \displaystyle I=\int  \frac { dt }{ t } =\log  t=\log  \left( 1+\sin ^{ 2 } x \right) $$
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Single Correct Medium Published on 17th 09, 2020
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