
reduction formula
i have shown
if
$\displaystyle
I_n=\int \frac {sin (2n\theta)}{sin \theta}
$
then
$\displaystyle
I_n=I_{n1}+ \frac {2}{2n1} sin (2n1)\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?

A few questions on and i have to prove a reduction formulae for sin (nx)/sin x
which will kill my problem. But i am still at loss of how to do it with the formula i have derived in part(a)