What is wrong with my integral? $\sin^5 x\cos^3 x$ I am trying to do an integration problem but am running into a problem! My answer is different from what the solution says. My attempt:

Evaluate $$\int\cos^3x\sin^5 x\mathop{dx}$$

$\int\cos^3x\sin^5 x\mathop{dx}=(1-\cos^2 x)^2\sin x \cos ^3 x$
Then with u substitution, letting $u=\cos x \implies du=-\sin x \mathop{dx}$
which gives us
$-\int (1-u^2)^2\cdot u^3\mathop{du}=\int-u^3+2u^5+u^7=-(\cos^ 4 x)/4+(\cos^6 x)/3-(\cos ^8 x)/8 +C$
However the solution books says the answer is 
$(\sin^6 x)/6-(\sin ^8 x)/8 +C$

Let's see a simpler case:
$$
\int 2\sin x\cos x\,dx
$$
You have two choices: either do $u=\sin x$, so $\cos x\,dx=du$, and you get
$$
\int 2u\,du=u^2+c=\sin^2x+c
$$
or do $v=\cos x$, so $\sin x\,dx=-dv$, and you get
$$
\int -2v\,dv=-v^2+c=-\cos^2x+c
$$
Which one is right? Both, of course, but this doesn't mean you reached the false conclusion that $\sin^2x=-\cos^2x$.
The fact is that an antiderivative is only determined up to a constant, so what you can say is that
$$
\sin^2x=-\cos^2x + k
$$
for some constant $k$; you surely know that $k=1$, in this case.
Your problem is exactly the same. You happened to use the second substitution instead of the first one. Try and determine $k$ such that
$$
-(\cos^ 4 x)/4+(\cos^6 x)/3-(\cos ^8 x)/8=(\sin^6 x)/6-(\sin ^8 x)/8 +k
$$
Hint: evaluate at $0$ both expressions.
