Question

Solve:$\int \frac{sinx.\cos x}{a{\cos }^{2}x+b{\sin }^{2}x}.dx$

Answer 1

Given the integral,

$\int \frac{\cos x\sin x}{a{\cos }^{2}x+b{\sin }^{2}x}dx$

Let us assume,

$u=b{\sin }^{2}x+a{\cos }^{2}x\Rightarrow \frac{du}{dx}=2b\cos x\sin x-2a\cos x\sin x\Rightarrow \frac{du}{dx}=\cos x\sin x(2b-2a)\Rightarrow \cos x\sin xdx=\frac{1}{2b-2a}du$

Substituting these values in the given integral we get,

$\int \frac{\cos x\sin x}{a{\cos }^{2}x+b{\sin }^{2}x}dx=\int \frac{1}{(2b-2a)u}du=\frac{1}{2b-2a}\int \frac{1}{u}du=\frac{1}{2b-2a}\ln \left(u\right)=\frac{\ln \left(u\right)}{2b-2a}=\frac{\ln (a{\cos }^{2}x+b{\sin }^{2}x)}{2b-2a}\therefore \int \frac{\cos x\sin x}{a{\cos }^{2}x+b{\sin }^{2}x}dx=\frac{\ln \left(|a{\cos }^{2}x+b{\sin }^{2}x|\right)}{2b-2a}+C.$