Question

Integrate:

$\int \frac{(3\sin x-2)\cos x}{(5-3\cos 2x-8\sin x)}dx$

• A. $I=\frac{1}{4}\ln (5-\cos 2x-8\sin x)+C$
• B. $I=\frac{1}{4}\ln (5-3\cos 2x-8\sin x)+C$
• C. $I=\frac{1}{4}\ln (2-3\cos x-8\sin x)+C$
• D. $I=\frac{1}{4}\ln (5-3\cos x-8\sin x)+C$

Answer 1

Answer: (B)

$I=\frac{1}{4}\ln (5-3\cos 2x-8\sin x)+C$

Consider the given integral.

$I=\int \frac{(3\sin x-2)\cos x}{(5-3\cos 2x-8\sin x)}dx$

Let $t=5-3\cos 2x-8\sin x$

$\frac{dt}{dx}=0-3(-\sin 2x)2-8\cos x$

$\frac{dt}{dx}=6\sin 2x-8\cos x$

$\frac{dt}{dx}=12\sin x\cos x-8\cos x$

$\frac{dt}{4}=\cos x(3\sin x-2)dx$

Therefore,

$I=\frac{1}{4}\int \frac{dt}{\left(t\right)}$

$I=\frac{1}{4}\ln \left(t\right)+C$

On putting the value of $t$, we get

$I=\frac{1}{4}\ln (5-3\cos 2x-8\sin x)+C$

Hence, this is the answer.