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Question

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


A
ln(1cos2x)+C
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B
ln(1+sin2x)+C
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C
ln(1+cos2x)+C
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D
ln(11cos2x)+C
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Solution

The correct option is C $$-\ln (1+\cos^2x)+C$$
$$\int { \frac { \sin { 2x }  }{ 1+\cos ^{ 2 }{ x }  }  } dx$$
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 $$\int { \frac { \sin { 2x }  }{ 1+\cos ^{ 2 }{ x }  } dx } =\int { \frac { -dt }{ t } =-\int { \frac { dt }{ t }  }  } \\ =-\log { \left| t \right| +c } \\ =-\log { \left( 1+\cos ^{ 2 }{ x }  \right) +c } $$

Mathematics

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