Compute the integral $ \int \frac{\sin x + \cos x}{\sqrt{9 + 16 \sin{2x}}} \mathrm{dx}$

Question

How to compute this indefinite integral.

$$ \int \frac{\sin x + \cos x}{\sqrt{9 + 16 \sin{2x}}} \mathrm{dx}$$

Answer 1

Let $u=\sin x-\cos x$ so $du=(\cos x + \sin x)dx$ and $u^2=1-\sin 2x$. Your integral is then $$\int\frac{du}{\sqrt{25-16u^2}}=\frac{1}{4}\mathrm{arcsin}\frac{4u}{5}+C=\frac{1}{4}\mathrm{arcsin}\frac{4(\sin x-\cos x)}{5}+C.$$