Question

$\int \frac{dx}{\sin x.\sin (x+a)}$ is equal to

• A. $cosecaln\left|\frac{\sin x}{\sin (x+a)}\right|+C$
• B. $cosecaln\left|\frac{\sin (x+a)}{\sin x}\right|+C$
• C. $cosecaln\left|\frac{\sin (x+a)}{\sec x}\right|+C$
• D. $cosecaln\left|\frac{\sec x}{\sec (x+a)}\right|+C$

Answer 1

The correct option is A $cosecaln\left|\frac{\sin x}{\sin (x+a)}\right|+C$

$\int \frac{dx}{\sin x\cdot \sin (x+a)}$

$=\int \frac{dx}{\sin x(\sin x\cos a+\cos x\sin a)}$

$=\int \frac{dx}{{\sin }^{2}x(\cos a+\frac{\cos x}{\sin x}\sin a)}$

$=\int \frac{dx}{{\sin }^{2}x(\cos a+\frac{\cos x}{\sin x}\sin a)}$

$=\int \frac{cose{c}^{2}xdx}{(\cos a+\cot x\sin a)}$

Let $\cot x=t$ then $cose{c}^{2}xdx=-dt$

$=-\int \frac{dt}{\cos a+\sin at}$ ........... $(\because \int \frac{1}{x}dx=lnx+c)$

$=-\frac{1}{\sin a}log|\cos a+\sin at|+c$

$=-cosecalog|\cos a+\sin a\cot x|+c$

$=-cosecalog|\cos a+\frac{\sin a\times \cos x}{\sin x}|+c$

$=-cosecalog\left|\frac{\sin x\cos a+\sin a\cos x}{\sin x}\right|+c$

$=+cosecalog\left|\frac{\sin x}{\sin (x+a)}\right|+c$ ...................$(\because -logx=log\frac{1}{x})$.