Question

Solve:
$\int \frac{\cos x}{6+4\sin x-{\cos }^{2}x}dx.$

Answer 1

$I=\int \frac{\cos x}{6+4\sin x-{\cos }^{2}x}dx$

Let $\sin x=t$

then $\cos xdx=dt$

$\therefore I=\int \frac{1}{6+4t-(1-t^2)}dt$

$=\int \frac{1}{t^2+4t+5}dt$

$=\int \frac{1}{(t+1)(t+4)}dt$

$=\frac{1}{3}\int [\frac{1}{t+1}-\frac{1}{t+4}]dt$

$=\frac{1}{3}[\log |t+1|-\log |t+4|]+c$

$=\frac{1}{3}\log \left|\frac{t+1}{t+4}\right|+c$

$\frac{1}{3}\log \left|\frac{\sin x+1}{\sin x+4}\right|+c$