How can I find the indefinite integral of $\int\sin^3x \cos^3x dx $?

Question

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

Answer 1

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.

Answer 2

You are in the right direction. Write $$ \sin^3x \cos^3x=\sin x\sin^2x \cos^3x=\sin x\,(1-\cos^2x) \cos^3x. $$

Answer 3

Note that $\sin^3x\cos^3x=\sin^3x\cos x-\sin^5x\cos x$ and $$\int\sin^kx\cos x\mathrm{d}x=\left[\frac{\sin^{k+1}x}{k+1}\right].$$

Answer 4

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.

Answer 5

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 $$ \int \cos^3(x)\sin^3(x)dx=-\frac{\cos^4(x)\sin^2(x)}{6}+\frac{2}{6}\int\cos^3(x)\sin(x)dx. $$

Answer 6

$$\begin{align} \int \sin^3x \cos^3xdx & = \int\sin^3x(1-\sin^2x)\cos x dx \\ &  (\sin x = t , \cos x dx = dt)\\ & = \frac{\sin^4x}{4} - \frac{\sin^6x}{6} + C \end{align}$$

$$\begin{align} \int \sin^3x \cos^3xdx & = \int\cos^3x(1-\cos^2x)\sin x dx \\ &  (\cos x = t , \sin x dx = -dt)\\ & = -\frac{\cos^4x}{4} + \frac{\cos^6x}{6} + C \end{align}$$

We can easily check that both the expressions differ by a constant by expanding one of them which can be adjusted with the $C$