Question

Evaluate

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

Answer 1

$\begin{array}{c}Wehave\\ \int \frac{3\cos x+2}{\sin x+2\cos x+3}dx\\ Here,\\ \left(3\cos x+2\right)=\lambda \left(\sin x+2\cos x+3\right)+\mu \left(\cos x-2\sin x\right)+u\\ \left(3\cos x+2\right)=\left(2\lambda +u\right)\cos x+\left(\lambda -2u\right)\sin x+\left(3\lambda +u\right)\\ \therefore 2\lambda +u=3,\lambda -2\mu =0,3\lambda +u=2\\ so,\\ \lambda =\frac{6}{5}\\ \mu =\frac{3}{5}\\ u=-\frac{8}{5}\\ I=\int \frac{6}{5}\frac{\left(\sin x+2\cos x+3\right)}{\left(\sin x+2\cos x+3\right)}dx+\frac{3}{5}\int \frac{\left(\cos x-2\sin x\right)dx}{\left(\sin x+2\cos x+3\right)}-\frac{8}{5}\int \frac{dx}{\sin x+2\cos x+3}\\ =\frac{6}{5}x+\frac{3}{5}\log \left(\sin x+2\cos x+3\right)-\frac{8}{5}{\tan }^{-1}\left(\frac{\tan \frac{x}{2}+1}{2}\right)+C\end{array}$