The value of $\int \frac{\sin x+\cos x}{3+\sin 2 x} d x$, is
Answer & explanation
Correct answer: option 1
We have,
$I =\int \frac{\sin x+\cos x}{3+\sin 2 x} d x$
$\Rightarrow I =\int \frac{1}{4-(1-\sin 2 x)} d(\sin x-\cos x)$
$\Rightarrow I =\int \frac{1}{2^2-(\sin x-\cos x)^2} d(\sin x-\cos x)$
$\Rightarrow I=\frac{1}{4} \log \left(\frac{2+\sin x-\cos x}{2-\sin x+\cos x}\right)+C$