Need help solving $\int\frac{\sin(x+1)\cos(x)}{\sin^2(x)+4} dx$

Question

As the title says I need help solving the indefinite integral

$$\int\frac{\sin(x+1)\cos(x)}{\sin^2(x)+4}dx$$

Thank you for any help.

Answer 1

Hint: $$\frac{\sin(x+1)\cos(x)}{\sin^2(x)+4} \,dx= \frac{(\sin{x}\cos{1}+\cos{x}\sin{1})\cos(x)}{\sin^2(x)+4} \,dx = \\ = \cos{1}\frac{\sin{x}\,d(\sin{x})}{\sin^2(x)+4}+ \sin{1}\frac{1-\sin^2{x}}{\sin^2(x)+4}\,dx.$$ Can you proceed?

Answer 2

Hint...start by expanding the $\sin(x+1)$ term using the compound angle identity, so thatvyou have two integrals $$\cos1\int\frac{\sin2x}{9-\cos 2x}dx+\sin1\int\frac{1}{1+4\tan^2x}dx$$

The first is immediately apparent, and the second will work out using $t=\tan x$ and application of partial fractions.