<s>[INST] Explanation of why $\int \sin^2 x\cos^2 x\; dx = 1/3 \sin^3 x - 2/5 \sin^5 x + 1/7 \sin^7 x +c$ 
verify the solution
  $$\int \sin^2 x\cos^5 x\; dx = 1/3 \sin^3 x - 2/5 \sin^5 x + 1/7 \sin^7 x +c$$

I have hit this in my book and can't work it out. Does anyone have any ideas or a walk-through that might help?
 [/INST] It's easier to start from the right hand side. Derive to get:
$$ (\sin^2 x - 2\sin^4 x + \sin^6 x)\cos x$$
Note that this is just:
$$(\sin x - \sin^3 x)^2\cos x=\sin^2 x(1-\sin^2x)^2\cos x=\sin^2 x\cos^5x$$
So, it seems you are missing a couple of powers of $\cos x$ in the integral..
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