Mathematics

$$ \int \dfrac{ \ sin2x}{1+ \ sinx} dx = $$


ANSWER

 $$ 2(1 + \sin x -log( 1+\ sin x) )+c$$


SOLUTION
$$\displaystyle \int{\dfrac{\sin 2x}{1+\sin x}dx}=\displaystyle \int{\dfrac{2\sin x.\cos x}{1+\sin x}dx}$$
Let $$1+\sin x=t$$
      $$\Rightarrow \cos x dx=dt$$
$$\Rightarrow \displaystyle \int{\dfrac{2\sin x\cos x}{1+\sin x}}dx=\displaystyle \int{\dfrac{2(t-1)}{t}}dt$$
                                       $$=2\displaystyle \int{\left(1-\dfrac{1}{t}\right)dt}$$
                                       $$=2\left[t-\log t\right]$$
                                       $$=2[1+\sin x-\log(1+\sin x)]$$
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Single Correct Medium Published on 17th 09, 2020
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