a/ Use intergration by parts to express:
in terms of I(n - 2)
b/ Hence, show that
It's a bit tricky, but here we go:
$\displaystyle I(n)= \int \sin^{n}(x) dx = \int \sin^{n-1}(x) \sin(x)dx$
Now we use integration by parts with $\displaystyle u=\sin^{n-1}(x), dv=\sin(x)dx $ and $\displaystyle du=(n-1)\sin^{n-2}(x)\cos(x)dx, v=-cos(x)dx$. So we get
$\displaystyle I(n)=-\sin^{n-1}(x)\cos(x)+(n-1)\int\sin^{n-2}(x)\cos^2(x)dx $
Using the identity $\displaystyle \sin^2(x)+\cos^2(x)=1$, we get
$\displaystyle I(n)=-\sin^{n-1}(x)\cos(x)+(n-1)\int\sin^{n-2}(x)(1-\sin^2(x))dx$
The integral on the RHS we recognize as $\displaystyle I(n-2)-I(n)$ so we get
$\displaystyle I(n) = -\sin^{n-1}(x)\cos(x) + (n-1)\left(I(n-2)-I(n)\right)$
Well the rest is very easy, just solve for $\displaystyle I(n) $ and you should have (a).
Putting it to the test for your given integral, we need $\displaystyle I(n-2)$ expressed in $\displaystyle I(n)$. Some basic manipulation should get you
$\displaystyle I(n-2) = \frac{n}{n-1}I(n) + \frac{\sin^{n-1}(x)\cos(x)}{n-1}$
Now your asked integral is basically $\displaystyle I(-4)=I(-2-2)$, with boundaries filled in. You should be able to do the rest on your own. I can confirm that the answer is indeed $\displaystyle \frac{4}{3}$.