How have I computed the integral $\int \sin^{3}(3x)\cos(3x)\,dx$ incorrectly? $\displaystyle \int \sin^{3}(3x)\cos(3x)\,dx $
$u = 3x  $
$du = 3\,dx  $
$3\displaystyle\int \sin^{3}(u)\cos(u)\,du$
$3\displaystyle\int \sin^{2}(u)\sin(u)\cos(u)\,du$
$3\displaystyle\int (1 - \cos^{2}(u))\sin(u)\cos(u)\,du$
$v = \cos(u)$
$dv = -\sin(u)\,du  $
$3\displaystyle\int (1 - v^{2})(-1)v\,dv$
$ 3\displaystyle\int (-v + v^{3})\,dv$
$ 3\left[-\frac{v^{2}}{2} + \frac{v^{4}}{4}\right]$
$3\left[ -\frac{\cos^{3}{3x}}{2} + \frac{\cos^{4}(3x)}{4}\right] $
$= -3 \frac{\cos^{2}(3x)}{2} + 3\frac{\cos^{4}(3x)}{4} + C$
wolfram states this as incorrect, what have I done wrong?
Thank you
 A: You determined $du=3\,dx$, but substituted $dx = 3\,du$.  So your answer is off by a factor of nine.
$$
\int \sin^3(3x) \cos(3x)\,dx = \int \sin^3(u) \cos(u)\cdot \frac{1}{3}\,du
= \frac{1}{3}\int \sin^3(u) \cos(u)\,du
$$
Then proceed as you did.  You'll get 
$$
\int \sin^3(3x) \cos(3x)\,dx =- \frac{1}{6} \cos^2(3x) -\frac{1}{12}\cos^4(3x)+C
$$
The other answers are suggesting substituting $u=\sin(3x)$ in one fell swoop.  Then $du = 3 \cos(3x)\,dx$, so
$$
\int \sin^3(3x) \cos(3x)\,dx = \frac{1}{3}\int u^3\,du = \frac{1}{3}\cdot\frac{1}{4}u^4 + C = \frac{1}{12}\sin^4(3x)+C
$$
Both are correct, although the second solution is more compact and concise.  In the comments you asked if the two are “equivalent.” Yes, but only in the antiderivative sense.  Using the trig identities:
\begin{align*}
    - \frac{1}{6}\cos^2(3x) + \frac{1}{12}\cos^4(3x)
    &= \frac{1}{12}\cos^2(3x)\left(\cos^2(3x) - 2\right) \\
    &= \frac{1}{12}\left(1-\sin^2(3x)\right)\left(1-\sin^2(3x)-2\right) \\
    &= \frac{1}{12}\left(1-\sin^2(3x)\right)\left(-1-\sin^2(3x)\right) \\
    &= \frac{1}{12}\left(\sin^2(3x)-1\right)\left(\sin^2(3x)+1\right) \\
    &= \frac{1}{12}\left(\sin^4(3x)-1\right) = \frac{1}{12}\sin^4(3x) - \frac{1}{12}
\end{align*}
You can see the two antiderivatives differ by a constant.

A: HINT:
Why are you making things complicated unnecessarily
Just set  $\sin ax=u$  for $$\int\sin^max\cos^{2n+1}ax\ dx$$
A: Your mistake is $dx = \frac 13 du$. So instead multiplying by $3$, multiply by $\frac13$.
Alternative -
I think if you put $\cos 3x$ as $u$. That is much easy to solve.
