i have shown

if

$\displaystyle

I_n=\int \frac {sin (2n\theta)}{sin \theta}

$

then

$\displaystyle

I_n=I_{n-1}+ \frac {2}{2n-1} sin (2n-1)\theta

$

the next part is hence,find

$\displaystyle

I_n=\int ^{\frac{\pi}{2}}_0 \frac {sin (5\theta)}{sin \theta}

$

Im missing something here. I split $\displaystyle sin(5\theta) $ as $\displaystyle sin(4\theta+\theta)$ etc

but all this does is evaluate the integral in terms of a reduction formula, not using part (a)

What is the trick i am missing to use part (a) and not just evaluate the second integral?