Question

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

• A. $\frac{1}{2}\log (3\cos x+2\sin x)+c$
• B. $\frac{1}{2}\log (3\cos x-2\sin x)+c$
• C. $\log (3\cos x+2\sin x)+c$
• D. $\log (3\cos x-2\sin x)+c$

Answer 1

Answer: (A)

$\frac{1}{2}\log (3\cos x+2\sin x)+c$

$\int \frac{2\cos x-3\sin x}{6\cos x+4\sin x}dx=\frac{1}{2}\int \frac{2\cos x-3\sin x}{3\cos x+2\sin x}dx$

Let $t=2\sin x+3\cos x\Rightarrow dt=(2\cos x-3\sin x)dx$

So, above integral becomes

$\frac{1}{2}\int \frac{dt}{t}=\frac{1}{2}\log t+c$

$=\frac{1}{2}\log (3\cos x+2\sin x)+c$