# Math Help - Integration by Parts Help Please :( Tough Question

1. ## Integration by Parts Help Please :( Tough Question

a/ Use intergration by parts to express:

in terms of I(n - 2)

b/ Hence, show that

2. It's a bit tricky, but here we go:

$I(n)= \int \sin^{n}(x) dx = \int \sin^{n-1}(x) \sin(x)dx$

Now we use integration by parts with $u=\sin^{n-1}(x), dv=\sin(x)dx$ and $du=(n-1)\sin^{n-2}(x)\cos(x)dx, v=-cos(x)dx$. So we get

$I(n)=-\sin^{n-1}(x)\cos(x)+(n-1)\int\sin^{n-2}(x)\cos^2(x)dx$

Using the identity $\sin^2(x)+\cos^2(x)=1$, we get

$I(n)=-\sin^{n-1}(x)\cos(x)+(n-1)\int\sin^{n-2}(x)(1-\sin^2(x))dx$

The integral on the RHS we recognize as $I(n-2)-I(n)$ so we get

$I(n) = -\sin^{n-1}(x)\cos(x) + (n-1)\left(I(n-2)-I(n)\right)$

Well the rest is very easy, just solve for $I(n)$ and you should have (a).
Putting it to the test for your given integral, we need $I(n-2)$ expressed in $I(n)$. Some basic manipulation should get you

$I(n-2) = \frac{n}{n-1}I(n) + \frac{\sin^{n-1}(x)\cos(x)}{n-1}$

Now your asked integral is basically $I(-4)=I(-2-2)$, with boundaries filled in. You should be able to do the rest on your own. I can confirm that the answer is indeed $\frac{4}{3}$.