Mathematics

$$\int \dfrac {x+\sin }{1+\cos x}dx=$$


ANSWER

$$x\ \tan \dfrac {x}{2}+c$$


SOLUTION
$$\displaystyle \int \dfrac {x+\sin x}{(1+\cos x)}dx=\displaystyle \int \dfrac {x+2 \sin (x/2)\cos (x/2)}{1+2\cos^2 (x/2)-1}dx$$
$$\displaystyle \int \dfrac {x+2\sin \left (\dfrac {x}{2}\right) \cos \dfrac {x}{2}}{2\cos^2x/2}dx =\displaystyle \int \dfrac {xdx}{2\cos^2 (x/2)}+\displaystyle \int tan \dfrac {x}{2}dx$$
$$\dfrac { 1 }{ 2 } \int { x } \sec ^{ 2 }{ \left( \dfrac { x }{ 2 }  \right) dx } +\int { \tan { \left( \dfrac { x }{ 2 }  \right)  } dx } $$
integrate first integral by parts
$$U=x$$
$$du=dx$$
$$du=\dfrac {1}{2}\sec^2 x/2 dx$$
$$u=\tan x/2$$
$$\dfrac {1}{2} \displaystyle \int x\sec^2 \left (\dfrac {x}{2}\right)dx +\displaystyle \int \tan \dfrac {x}{2}dx =x \tan \dfrac {x}{2}-\displaystyle \int \tan \dfrac {x}{2}dx +\displaystyle \int \tan \dfrac {x}{2}dx$$
$$=x\tan (x/2)+c$$
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Single Correct Medium Published on 17th 09, 2020
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Assume: $$\displaystyle I=\int \frac{\sqrt{\cos 2x}}{\sin x}dx$$

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