Mathematics

# $\displaystyle \int \dfrac {\sin 2x}{1+\cos^2x}dx$

$-\ln (1+\cos^2x)+C$

##### SOLUTION
$\int { \frac { \sin { 2x } }{ 1+\cos ^{ 2 }{ x } } } dx$
Put $1+\cos ^{ 2 }{ x } =t\\ 0+2\cos { \left( -\sin { x } \right) dx=dt } \\ -2\sin { x } \cos { x } dx=dt\\ -\sin { 2x } dx=dt\\ \sin { 2xdx=-dt }$
Thus $\int { \frac { \sin { 2x } }{ 1+\cos ^{ 2 }{ x } } dx } =\int { \frac { -dt }{ t } =-\int { \frac { dt }{ t } } } \\ =-\log { \left| t \right| +c } \\ =-\log { \left( 1+\cos ^{ 2 }{ x } \right) +c }$

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Single Correct Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84

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