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


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 $f(x)=g(x)+C, f'(x)=g'(x)$ 
