<s>[INST] 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?
 [/INST] 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.
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