Question

$\int \frac{sin2x}{sin5xsin3x}dx$

• A. $lnsin3x-lnsin5x+c$
• B. $\frac{1}{3}lnsin3x+\frac{1}{5}lnsin5x+c$
• C. $\frac{1}{3}lnsin3x-\frac{1}{5}lnsin5x+c$
• D. $3lnsin3x-5lnsin5x+c$

Answer 1

The correct option is C $\frac{1}{3}lnsin3x-\frac{1}{5}lnsin5x+c$

$\int \frac{\sin 2x}{\sin 5x\sin 3x}dx$

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

$=\int \frac{\sin 5x\cos 3x-\cos 5x\sin 3x}{\sin 5x\sin 3x}dx$ using $\sin (A-B)=\sin A\cos B-\cos A\sin B$

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

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

$=\int \cot 3xdx-\int \cot 5xdx$

$=\frac{1}{3}\ln \sin 3x-\frac{1}{5}\ln \sin 5x+c$ using the formula $\int \cot axdx=\frac{1}{a}\ln \sin 3x$

where $c$ is the constant of integration