<s>[INST] How can I find the indefinite integral of $\int\sin^3x \cos^3x dx $? I am looking to evaluate the indefinite integral
\begin{equation*}
\int\sin^3x \cos^3x dx. 
\end{equation*}
I'm not sure if I started this right but I broke the terms up like this:
\begin{equation*}
\int\sin^2x \sin x \cos^2x \cos x 
\end{equation*}
Edit: I got 
\begin{equation*}
{1 \over 4}\cos^4x + {1 \over 6}\cos^6x + C 
\end{equation*}
but my book says it should be sine not cosine
 [/INST] $$\begin{align}
\int \sin^3x \cos^3xdx
& = \int\sin^3x(1-\sin^2x)\cos x dx \\ 
&  (\sin x = t , \cos x dx = dt)\\
& = \frac{\sin^4x}{4} - \frac{\sin^6x}{6} + C
\end{align}$$
$$\begin{align}
\int \sin^3x \cos^3xdx
& = \int\cos^3x(1-\cos^2x)\sin x dx \\ 
&  (\cos x = t , \sin x dx = -dt)\\
& = -\frac{\cos^4x}{4} + \frac{\cos^6x}{6} + C
\end{align}$$
We can easily check that both the expressions differ by a constant by expanding one of them which can be adjusted with the $C$
</s>