Question

Integrate

$\int \frac{\cos x-\sin x}{7-9\sin 2x}dx$

Answer 1

$\int \frac{\cos x-\sin x}{7-9\sin 2x}dx$

$=\int \frac{\cos x-\sin x}{16-9(1+\sin 2x)}dx$

$=\int \frac{\cos x-\sin x}{16-9(\sin x+\cos x{)}^2}dx$

substitute $\sin x+\cos x=t\to (\cos x-\sin x)dx=dt$

$=\int \frac{dt}{16-9t^2}$

$=\frac{1}{24}\left[\log \left|\frac{4-3t}{4+3t}\right|\right]+C$

$=\frac{1}{24}\left[\log \left|\frac{4-3(\sin x+\cos x)}{4+3(\sin x+\cos x)}\right|\right]+C$