Integrals — Class 12 Maths Solution

.: Let $I = \int {\cfrac{{2\cos x - 3\sin x}}{{6\cos x + 4\sin x}}dx = \cfrac{1}{2}} \int {\cfrac{{2\cos x - 3\sin x}}{{2\sin x + 3\cos x}}dx}$

Put $2{\mathop{\rm sinx}\nolimits} + 3cosx = t$ $\Rightarrow$ $\left( {2\cos x - 3\sin x} \right)dx = dt$

$\therefore$ $I = \cfrac{1}{2}\int {\cfrac{{dt}}{t} = \cfrac{1}{2}\log \left| t \right| + C = \cfrac{1}{2}\log \left| {2\sin x + 3\cos x} \right| + C}$