Question

If $\int \frac{\sin 2x}{\sin 5x\sin 3x}dx=\frac{1}{3}\log |\sin 3x|-\frac{1}{5}\log |f\left(x\right)|+C$, then $f\left(x\right)=$ _________.

• A. $\sin 5x$
• B. $\sin 4x$
• C. $\sin 2x$
• D. $\sin 6x$

Answer 1

Answer: (A)

$\sin 5x$

Given $\int \frac{\sin 2x}{\sin 5x\sin 3x}dx=\frac{1}{3}\log |\sin 3x|-\frac{1}{5}\log |f\left(x\right)|+C$ ...... $\left(i\right)$

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

$=\int \frac{\sin 5x\cos 3x-\cos 5x\sin 3x}{\sin 5x\sin 3x}dx$

$=\int (\cot 3x-\cot 5x)dx$

$=\frac{1}{3}\log |\sin 3x|-\frac{1}{5}\log |\sin 5x|+C$

COmpairing with equation $\left(i\right)$, we get
$f\left(x\right)=\sin 5x$

Hence, option A is correct.