# Thread: Integration - Reduction formulas

1. ## Integration - Reduction formulas

By writing $\sin^{2n} (x) = \sin^{2n-2}( x ) [1- \cos^2( x )]$, use integration by parts (just once, in the second term) to derive the reccurence relation:

$I_{2n} = \left(\frac{2n-1}{2n}\right)I_{2n-2}$ $(n \ge 1)$, for $I_{2n} = \int_{0}^{\pi} \sin^{2n}( x )dx$.

Hence find the value of $I_4$ without using your calculator at any stage.

2. Originally Posted by mitch_nufc
By writing $sin^{2n}( x ) = [sin^{2n-2}( x )][1-cos^2( x )]$, use integration by parts (just once, in the second term) to derive the reccurence relation:
$I_{2n} = (\frac{2n-1}{2n})I_{2n-2}$ $(n \ge 1)$, for $I_{2n} = \int_{0}^{\pi}sin^{2n}( x )dx$.
Hence find the value of $I_4$ without using your calculator at any stage.
$I_{2n} = \int_{0}^{\pi}sin^{2n}( x )dx$

$= \int_{0}^{\pi}[sin^{2n-2}( x )][1-cos^2( x )]dx$

$= \int_{0}^{\pi}sin^{2n-2}( x )-sin^{2n-2}cos^2( x )dx$

Leave the first term as it is, since you're trying to find a solution in terms of $\int_{0}^{\pi}sin^{2n-2}(x)$, since $I_{2n-2} = \int_{0}^{\pi} sin^{2n-2}(x)$

So the integral is now:

$I_{2n} = I_{2n-2}-\int_{0}^{\pi} sin^{2n-2}(x)cos^2( x )dx$

Carry out integration by parts on the 2nd term of the integral and again try to manipulate it so that you get an expression in terms of $\int_{0}^{\pi}sin^{2n-2}(x)$. You should then be able to factor out $\int_{0}^{\pi}sin^{2n-2}(x)$ of the whole expression, and the multiple should be $(\frac{2n-1}{2n})$.

Hint: When it comes to integrating that term by parts, it is difficult to integrate $\int_{0}^{\pi}sin^{2n-2}( x )$, hence that is the term you should differentiate!

Hint #2: $\int cos^2(x) = \frac{x}{2} + \frac{sin(2x)}{4} +C$