Question

The value of $\int \frac{\sin x+\cos x}{3+\sin 2x}dx$ is

• A. $\frac{1}{4}\log \left(\frac{2-\sin x+\cos x}{2+\sin x-\cos x}\right)+C$
• B. $\frac{1}{2}\log \left(\frac{2+\sin x}{2-\sin x}\right)+C$
• C. $\frac{1}{4}\log \left(\frac{1+\sin x}{1-\sin x}\right)+C$
• D. None of the above

Answer 1

The correct option is A $\frac{1}{4}\log \left(\frac{2-\sin x+\cos x}{2+\sin x-\cos x}\right)+C$

Let $I=\int \frac{\sin x+\cos x}{3+\sin 2x}dx$
$=\int \frac{\sin x+\cos x}{4+\sin 2x-1}dx$
$\Rightarrow I=\int \frac{\sin x+\cos x}{4-{(\sin x-\cos x)}^2}dx$
Put $\sin x-\cos x=t$
$\Rightarrow (\sin x+\cos x)dx=dt$
Therefore, $I=\int \frac{dt}{4-t^2}=\frac{1}{4}\log \frac{2-t}{2+t}+C$
$=\frac{1}{4}\log \left(\frac{2-\sin x+\cos x}{2+\sin x-\cos x}\right)+C$