Question

$\int \frac{3sinx+2\cos x}{3\cos x+2\sin x}.dx$

Answer 1

Let, $I=\int \frac{3\sin x+2\cos x}{3\cos x+2\sin x}dx$

We write,

$3\sin x+2\cos x=\lambda (3\cos x+2\sin x)+\mu \left[\frac{d}{dx}\right(3\cos x+2\sin x\left)\right]=\lambda (3\cos x+2\sin x)+\mu (-3\sin x+2\cos x)=\sin x(2\lambda -3\mu )+\cos x(3\lambda +2\mu )$

Thus, $2\lambda -3\mu =3⟶\left(1\right)3\lambda +2\mu =2⟶\left(2\right)$

Solving equation $\left(1\right)$ and $\left(2\right)$ we get,

$\lambda =\frac{12}{13}\mu =\frac{-5}{13}$

$\therefore I=\int \frac{\mu (-3\sin x+2\cos x)+\lambda (3\cos x+2\sin x)}{3\cos x+2\sin x}dx=\lambda \int dx+\mu \int \frac{-3\sin x+2\cos x}{3\cos x+2\sin x}dx$

Now, let $3\cos x+2\sin x=t\Rightarrow (-3\sin x+2\cos x)dx=dt\therefore \int \frac{-3\sin x+2\cos x}{3\cos x+2\sin x}dx=\int \frac{dt}{t}=\log \left|t\right|+C=\log (3\cos x+2\sin x)+C$

putting the value in $I$,

$I=\lambda \int dx+\mu \int \frac{-3\sin x+2\cos x}{3\cos x+2\sin x}dx=\lambda x+\mu \left[\log (3\cos x+2\sin x)\right]+C=\frac{12}{13}x-\frac{5}{13}\log (3\cos x+2\sin x)+C$

$\therefore \int \frac{3\sin x+2\cos x}{3\cos x+2\sin x}dx=\frac{12}{13}x-\frac{5}{13}\log (3\cos x+2\sin x)+C$