Question

If $\int \frac{3+2\cos x}{(2+3\cos x{)}^2}dx=\frac{f\left(x\right)}{2+3\cos x}+C$, where $C$ is a constant of integration, then $f\left(x\right)$ is

• A. $\cos x$
• B. $2+3\cos x$
• C. $\sin x$
• D. $3+2\sin x$

Answer 1

The correct option is C $\sin x$

$I=\int \frac{3+2\cos x}{(2+3\cos x{)}^2}dx$
Divide both the numerator and the denominator by ${\sin }^{2}x$
$I=\int \frac{(3{\text{cosec}}^{2}x+2\cot x\text{cosec}x)}{(2\text{cosec}x+3\cot x{)}^2}dx$

Put $2\text{cosec}x+3\cot x=t$
$\Rightarrow -(3{\text{cosec}}^{2}x+2\cot x\text{cosec}x)dx=dt$
$\therefore I=\int \frac{-1}{t^2}dt=\frac{1}{t}+C=\frac{1}{2\text{cosec}x+3\cot x}+C=\frac{\sin x}{2+3\cos x}+C$

Hence, $f\left(x\right)=\sin x$