High School Trigonometric Integration I am wondering which step has gone wrong. Is it wrong to use $u=\sin x$ ?
$$ \int \cos^3x\ \sin x\;\mathrm{d}x $$
$$=\int\cos^2x \sin x \cos x\;\mathrm{d}x $$
$$=\int(1-\sin^2x)\sin x\;\mathrm{d}(\sin x)$$
$$=\int\sin x\;\mathrm{d}(\sin x)-\int\sin^3x\;\mathrm{d}(\sin x)$$
$$=\frac 12 \sin^2x-\frac14\sin^4x+C$$
 A: It's fine. You should get an answer that differs from the answer you'd get substituting $u=\cos x$ instead. What's the explanation? The two answers differ by a constant.
But try your method with $\int \cos^2x\sin x\,dx$. Good luck!!

To be a little bit more explicit:
Since $\cos^2 x + \sin^2 x = 1$ you have
$$ 1 = (\cos^2 x + \sin^2 x)^2 = \cos^4 x + 2 \sin^2 x \cos^2 x + \sin^4 x $$
Using the identity again on the middle term you have 
$$ 1 = \cos^4 x - \sin^4 x + 2 \sin^2 x $$
and so the answer you get from the $\sin x$ substitution differ form the $\cos x$ substitution by exactly a constant ($\frac14$ in fact). 
A: no you are not wrong but this can also be done without substitution,
recall your high school formula: $\int x^n\ dx=\frac{x^{n+1}}{n+1}+C$  $$\int \cos^3x\sin x\ dx=-\int (\cos x)^3\ d(\cos x)=-\frac{(\cos x)^4}{4}+C$$
A: Your work is correct.  You can check your work by differentiating your result:
\begin{align*}
\frac{d}{dx}\left(\frac 12 \sin^2x-\frac14\sin^4x+C\right)&=\sin x\cos x-\sin^3 x\cos x \\
&=\left(1-\sin^2 x\right)\sin x \cos x \\
&=\sin x\cos^3x
\end{align*}
This is the original integrand, so your integration was correct.
A: You can work out such integrals using the complex definition of the trigonometric functions.
With $z=e^{ix}$ (and $dz=ie^{ix}dx$), $\cos x:=\dfrac{z+z^{-1}}2,\sin x=\dfrac{z-z^{-1}}{2i}$ we have
$$ \int \cos^3x\ \sin x\,dx=\int\left(\dfrac{z+z^{-1}}2\right)^3\dfrac{z-z^{-1}}{2i}\frac{dz}{iz}=-\int\frac{z^3+2z-2z^{-3}-z^{-5}}{16}dz\\
=-\frac{\dfrac{z^4+z^{-4}}4+z^2+z^{-2}}{16}=-\frac1{32}\cos4x-\frac18\cos 2x.$$
