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
 A: Let's "linearise" the integrand
$$\begin{align}\sin^3{x}\cos^3{x}&=\left(\frac{\sin{2x}}{2}\right)^3\\
&=\frac{1}{8}\left(\frac{e^{2ix}-e^{-2ix}}{2i}\right)^3\\
&=\frac{1}{8}\frac{e^{6ix}-3e^{2ix}+3e^{-2ix}-e^{-6ix}}{-8i}\\
&=\frac{-\sin{6x}+3\sin{2x}}{32}\end{align}$$
And it boils down to integrating two sines.
A: You are in the right direction. Write
$$
\sin^3x \cos^3x=\sin x\sin^2x \cos^3x=\sin x\,(1-\cos^2x) \cos^3x.
$$
A: Note that $\sin^3x\cos^3x=\sin^3x\cos x-\sin^5x\cos x$ and
$$\int\sin^kx\cos x\mathrm{d}x=\left[\frac{\sin^{k+1}x}{k+1}\right].$$
A: Well, you can do the following:
$$ \int -\cos^{3}x (-1+\cos^{2}x) \sin x dx = \int (\sin x \cdot \cos^{3}xdx -\sin x \cdot \cos^{5}x )dx = \int sinx \cdot \cos^{3}x dx + \int -\sin x \cdot \cos^{5}x dx $$
Then you can easily solve that, by using a change of variable. Hope it helps.
A: $$\begin{align}
\int \sin^3x \cos^3xdx
& = \int\sin^3x(1-\sin^2x)\cos x dx \\ 
&  (\sin x = t , \cos x dx = dt)\\
& = \frac{\sin^4x}{4} - \frac{\sin^6x}{6} + C
\end{align}$$
$$\begin{align}
\int \sin^3x \cos^3xdx
& = \int\cos^3x(1-\cos^2x)\sin x dx \\ 
&  (\cos x = t , \sin x dx = -dt)\\
& = -\frac{\cos^4x}{4} + \frac{\cos^6x}{6} + C
\end{align}$$
We can easily check that both the expressions differ by a constant by expanding one of them which can be adjusted with the $C$
A: You might want to use some trigonometric identities here to simply the integrand
We know that $$\sin(2x)=2\sin(x)\cos(x)$$ $$\implies \frac{1}{8}\sin^3(2x)=\sin^3(x)\cos^3(x)$$ Replacing the integrand with its equivalent that we calculated. We get $$\int \sin^3(x)\cos^3(x)dx=\frac{1}{8}\int \sin^3(2x)dx$$ If it still seems difficult to evaluate you might want to do a U-Substitution. Writing $u=2x$, we get the following information $$\begin{pmatrix}u \\ du\end{pmatrix}=\begin{pmatrix}2x\\ 2 \ dx\end{pmatrix}$$ $$\implies \frac{1}{8}\int \sin^3(u)(\frac{du}{2})=\frac{1}{16}\int \sin^3(u)du$$ Can you proceed further by using power-reduction formulas for trigonometric functions. Cheers
A: The reduction formula 
\begin{equation*}
\int \cos^m(x)\sin^n(x)dx=-\frac{\cos^{m+1}(x)\sin^{n-1}(x)}{m+n}+\frac{n-1}{m+n}\int\cos^m(x)\sin^{-2+n}(x)dx
\end{equation*} 
with $m=n=3$ gives
$$
\int \cos^3(x)\sin^3(x)dx=-\frac{\cos^4(x)\sin^2(x)}{6}+\frac{2}{6}\int\cos^3(x)\sin(x)dx.
$$
