Question

If $\int \sin x\cdot \ln \left(\sin x\right)dx=a\cos x(1-\ln (\sin x\left)\right)+\ln |\tan \left(bx\right)|+C,$ then the value of $\frac{a}{b}$ is equal to

Answer 1

$I=\int \sin x\cdot \ln \left(\sin x\right)dx\underset{II}{\downarrow }\underset{I}{\downarrow }$
Applying integration by parts
$\Rightarrow I=\ln \left(\sin x\right)\cdot (-\cos x)-\int \frac{\cos x}{\sin x}\cdot (-\cos x)dx=-\ln \left(\sin x\right)\cdot \cos x+\int \frac{1-{\sin }^{2}x}{\sin x}dx=-\ln \left(\sin x\right)\cdot \cos x+\int \text{cosec}xdx-\int \sin xdx\Rightarrow I=-\ln \left(\sin x\right)\cdot \cos x+\ln \left|\tan \left(\frac{x}{2}\right)\right|+\cos x+C$
$\Rightarrow I=\cos x(1-\ln (\sin x\left)\right)+\ln \left|\tan \left(\frac{x}{2}\right)\right|+C$
$\Rightarrow a=1,b=\frac{1}{2}$
$\Rightarrow \frac{a}{b}=2$