Question

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

Answer 1

\[\int\frac{2\cos x - 3\sin x}{6\cos x + 4\sin x}dx\]
\[ \Rightarrow \int\frac{2\cos x - 3\sin x}{2\left( 3\cos x + 2\sin x \right)}dt\]
\[Let, 3\cos x + 2\sin x = t\]
\[ \Rightarrow 2\cos x - 3\sin x = \frac{dt}{dx}\]
\[ \Rightarrow \left( 2\cos x - 3\sin x \right)dx = dt\]
\[Now, \int\frac{2\cos x - 3\sin x}{2\left( 3\cos x + 2\sin x \right)}dt\]
\[ = \int\frac{dt}{2t}\]
\[ = \frac{1}{2}\text{log}\left| t \right| + C\]
\[ = \frac{1}{2} \text{log} \left| 3\cos x + 2\sin x \right| + C\]