Is $\int\frac{\cos^5x\sin^3x}{1+\cos2x}dx = \frac{\sin^4x}{8}-\frac{\sin^6x}{12} +C$ or $\frac{\cos^6x}{12}-\frac{\cos^4x}{8} +C$? $\int\dfrac{\cos ^5x\sin ^3x}{1+\cos 2x}dx = \dfrac{\sin ^4x}{8}-\dfrac{\sin ^6x}{12} +C$ or $\dfrac{\cos ^6x}{12}-\dfrac{\sin ^4x}{8} +C$?
$\int\dfrac{\cos ^5x\sin ^3x}{1+\cos 2x}dx$ = $\dfrac{1}{2}\int{\cos ^3x\sin ^3x} dx$
1) $\dfrac{1}{2}\int{\cos ^3x\sin ^3x} dx$ 
= $\dfrac{1}{2}\int{\cos ^2x\sin ^3x\cos x} dx$
= $\dfrac{1}{2}\int{(1-\sin ^2x)\sin ^3x} d\sin x$
= $\dfrac{1}{2}\int{\sin ^3x-\sin ^5x} d\sin x$
= $\dfrac{\sin ^4x}{8}-\dfrac{\sin ^6x}{12} +C$
2) $\dfrac{1}{2}\int{\cos ^3x\sin ^3x} dx$ 
= $\dfrac{1}{2}\int{\cos ^3x\sin ^2x\sin x} dx$
= $-\dfrac{1}{2}\int{(1-\cos ^2x)\cos ^3x} d\cos x$
= $-\dfrac{1}{2}\int{\cos ^3x-\cos ^5x} d\cos x$
= $\dfrac{\cos ^6x}{12}-\dfrac{\cos ^4x}{8} +C$
Which one is correct? or both are correct?

You can differentiate and simplify your two answers to see that they are both correct. If you take the first answer, expand it using the identity $\sin^2 x = 1-\cos^2 x$, and simplify, you will get the second answer plus an additional constant. Thus the two expressions (ignoring the $C$'s) differ by a constant.
