Question

Integration of sinx/cos3x cos2x is ???

Answer 1

We have to find the indefinite integral $\int {\frac{\sin x}{\cos (3x)\cos (2x)}} \, dx$.

Now $\sin x=\sin (3x-2x)\\ \sin x=\sin (3x)\cos (2x)-\cos (3x) \sin (2x)$

The integral becomes,

$\int {\frac{\sin x}{\cos (3x)\cos (2x)}} \, dx=\int {\frac{\sin (3x)\cos (2x)-\cos (3x) \sin (2x)}{\cos (3x)\cos (2x)}} \, dx\\ \int {\frac{\sin x}{\cos (3x)\cos (2x)}} \, dx=\int [{\tan (3x)-\tan(2x)] \, dx$

$\int {\frac{\sin x}{\cos (3x)\cos (2x)}} \, dx=\frac{1}{3} \ln|\sec(3x)|-\frac{1}{2} \ln|\sec(2x)|+C$