Mathematics

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

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

##### 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
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 86

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