Question

If $\int \frac{\sin x}{\sin \left(x-\alpha \right)}dx=Ax+B\log \sin \left(x-\alpha \right)+C,$ then value of $(A,B)$ is

• A. $\left(\sin \alpha ,\cos \alpha \right)$
• B. $\left(-\cos \alpha ,\sin \alpha \right)$
• C. $\left(-\sin \alpha ,\cos \alpha \right)$
• D. $\left(\cos \alpha ,\sin \alpha \right)$

Answer 1

The correct option is D $\left(\cos \alpha ,\sin \alpha \right)$

$\int \left(\frac{sinx}{sin(x-\alpha )}\right)dx$

$let(x-\alpha )=t$

$\therefore x=\alpha +t$

$dx=dt$

$I=\int \left(\frac{sin(\alpha +t)}{sin\left(t\right)}\right)dx$

$=\int \left(\frac{sin\alpha cost+cos\alpha sint}{sint}\right)dx$

$=\int sin\alpha \left(\frac{cost}{sint}\right)dt+\int cos\alpha dt$

$=sin\alpha \int \left(\frac{cost}{sint}\right)dt+cos\alpha \int dt$

$=sin\alpha log\left(sint\right)+cos\alpha \left(t\right)+C$

$=sin\alpha logsin(x-\alpha )+cos\alpha (x-\alpha )+C$

$=xcos\alpha +sin\alpha logsin(x-\alpha )+C_1$

Compare it with $Ax+Blogsin(x-\alpha )+C$

we get A=$cos\alpha andB=sin\alpha$