Question

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

• A. $logsin2x+c$
• B. $2(1+\sin x-log(1+sinx\left)\right)+c$
• C. $\frac{1}{2}log(1+si{n}^{2}x)+c$
• D. $ta{n}^{-1}\left(sinx\right)+c$

Answer 1

The correct option is B $2(1+\sin x-log(1+sinx\left)\right)+c$

$\int \frac{\sin 2x}{1+\sin x}dx=\int \frac{2\sin x.\cos x}{1+\sin x}dx$

Let $1+\sin x=t$

$\Rightarrow \cos xdx=dt$

$\Rightarrow \int \frac{2\sin x\cos x}{1+\sin x}dx=\int \frac{2(t-1)}{t}dt$

$=2\int (1-\frac{1}{t})dt$

$=2[t-\log t]$

$=2[1+\sin x-\log (1+\sin x\left)\right]$