Question

If $\int \frac{\sin x}{\sin (x-\alpha )}dx=Ax+B\log \sin (x-\alpha )+c$ then find the value of $(A,B)$

Answer 1

$\int \frac{\sin x}{\sin (x-\alpha )}dx=Ax+B\log \sin (x-\alpha )+c$

Differentiating w.r.t to $x$ both sides, we get

$\Rightarrow \frac{\sin x}{\sin (x-\alpha )}=A+\frac{B\cos (x-\alpha )}{\sin (x-\alpha )}$

$\Rightarrow \sin x=A\sin (x–\alpha )+B\cos (x–\alpha )$

$\sin x=A(\sin x\cos \alpha –\cos x\sin \alpha )+B(\cos x\cos \alpha +\sin x\sin \alpha )$

$sinx=sinx(Acos\alpha +Bsin\alpha )+cosx(Bcos\alpha –Asin\alpha )$

Now solving $Acos\alpha +Bsin\alpha =1$ and $Bcos\alpha –Asin\alpha =0$

$\Rightarrow B\cos \alpha =A\sin \alpha$

$\Rightarrow B=\frac{A\sin \alpha }{\cos \alpha }$

Substituting the value of $B=\frac{A\sin \alpha }{\cos \alpha }$ in $Acos\alpha +Bsin\alpha =1$ we get

$Acos\alpha +\frac{A\sin \alpha }{\cos \alpha }\times sin\alpha =1$

$A{\cos }^2\alpha +A{\sin }^2\alpha =\cos \alpha$

$\Rightarrow A({\cos }^2\alpha +{\sin }^2\alpha )=\cos \alpha$

$\therefore A=\cos \alpha$

Substitute $A=\cos \alpha$ in $B\cos \alpha -A\sin \alpha =0$

$\Rightarrow B\cos \alpha -\cos \alpha \times \sin \alpha =0$

$\Rightarrow B\cos \alpha =\cos \alpha \times \sin \alpha$

$\therefore B=\sin \alpha$

Hence $(A,B)=(\cos \alpha ,\sin \alpha )$