Mathematics

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

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

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

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