Question

$\int \frac{2\cos x-3\sin x}{6\cos x+4\sin x}dx$

Answer 1

$$\begin{gathered} \int \frac{2\cos x-3\sin x}{6\cos x+4\sin x}dx \\ \Rightarrow \int \frac{2\cos x-3\sin x}{2\left(3\cos x+2\sin x\right)}dt \\ \mathrm{Let},3\cos x+2\sin x=t \\ \Rightarrow 2\cos x-3\sin x=\frac{dt}{dx} \\ \Rightarrow \left(2\cos x-3\sin x\right)dx=dt \\ \mathrm{Now},\int \frac{2\cos \mathrm{x}-3\sin \mathrm{x}}{2\left(3\cos \mathrm{x}+2\sin \mathrm{x}\right)}\mathrm{d}\mathrm{t} \\ =\int \frac{dt}{2t} \\ =\frac{1}{2}\log \left|t\right|+C \\ =\frac{1}{2}\log \left|3\cos x+2\sin x\right|+C \end{gathered}$$