Integration giving different answers (trig substitution) Integrating $\sin^3x\cos^5x$, i get 2 different answers, using techniques that should both be valid.

 A: The one on the left has a sign error. Presumably you had let $w=\cos\theta$. Then $dw=-\sin\theta\,d\theta$.  So the answer on the left is off by a factor of $-1$.
After that is corrected, the answers will still look different. That is taken care of by the constant of integration. This sort of thing happens a lot, particularly with trigonometric functions, because of the many identities.
As a simpler example, both $-\cos^2 t+C$ and $\sin^2 t+C$ are correct answers to $\int 2\sin t\cos t\,dt$.
A: HINT:
Establish that $$\frac{\cos^8x-\sin^8x}8-\frac{\cos^6x-\sin^6x}6-\frac{\sin^4x}4$$ is a constant
If $u_n=\cos^nx-\sin^nx,$
$$(\cos^nx-\sin^nx)(\cos^2x+\sin^2x)=\cdots$$
$$\implies u_n=u_{n+2}+\cos^2x\sin^2x\cdot u_{n-2}$$
$$\implies u_{n+2}=u_n-\cos^2x\sin^2x\cdot u_{n-2}$$
Now, $u_0=0,u_2=\cos^2x-\sin^2x$
$$n=2\implies u_4=u_2-\cos^2x\sin^2x\cdot u_0=u_2$$
$$n=4\implies u_6=u_4-\cos^2x\sin^2x\cdot u_2=u_2(1-\cos^2x\sin^2x)$$
$$n=6\implies u_8=u_6-\cos^2x\sin^2x\cdot u_4=u_2(1-\cos^2x\sin^2x)-u_2=-u_2(\cos^2x\sin^2x)$$
If $f(x)=g(x)+C, f'(x)=g'(x)$ 
