Question

Evalaute

$\int \frac{\cos xdx}{6+4\sin x-{\cos }^{2}x}$

Answer 1

$wehave$

$\int \frac{\cos xdx}{6+4\sin x-{\cos }^{2}x}$

$=\int \frac{\cos xdx}{6+4\sin x-1+{\sin }^{2}x}$

$=\int \frac{\cos xdx}{{\sin }^{2}x+4\sin x+5}$

$=\int \frac{\cos xdx}{{\left(\sin x+2\right)}^2+1}$

$put\sin x=t$

$\cos xdx=dt$

$=\int \frac{dt}{{\left(t+2\right)}^2+1}$

$={\tan }^{-1}\left(t+2\right)+C$

$={\tan }^{-1}\left(\sin x+2\right)+C$