Proving a trigonometric identity by integration. By considering, $$\int \sin^n (x)\cos^3(x)dx$$
Prove that: 
$$ \frac{ \sin^8(x)}{8} - \frac{ \sin^6(x)}{6} + \frac{1}{24} =
   \frac{\cos^8(x)}{8} - \frac{ \cos^6(x)}{3} + \frac{ \cos^4(x)}{4}$$
I have integrated the suggested integral successfully, that being:
$$ \frac{\sin^{n+1}(x)}{n+1} - \frac { \sin^{n+3}(x)}{n+3} +c $$
I see some resemblance but I fail to connect and finish the proof. Any hints on finding the relation between the powers of sin and cos through this integral? Obviously we could expand $\sin^2(x) = (1-\cos^2(x))$ but this seems far too tedious. 

Hint:
$$I=\int\sin^{2m+1}x\cos^{2n+1}x\ dx=\int(1-\cos^2x)^m\cos^{2n+1}x\sin x dx$$
Set $\cos x=u$
Again,
$$I=\int(1-\sin^2x)^n\sin^{2m+1}x\cos x\ dx$$
Set $\sin x=v$
Here $2n+1=3$
Set $2m+2n+2=8,6,4$ etc.
