reduction formula

• February 28th 2010, 01:04 PM
jiboom
reduction formula
i have shown

if

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

then

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

the next part is hence,find

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

Im missing something here. I split $sin(5\theta)$ as $sin(4\theta+\theta)$ 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?
• March 1st 2010, 10:09 AM
jiboom
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)