# Integrating $\int \sin^3(a x)dx$

I have to integrate the following, where a is a constant:

$$I=\int \sin^3(a x)dx$$

I did the following: $$u=ax$$ $$\frac{du}{a}=dx$$

Which gets me to this point:

$$\frac{1}{a}\int(1-\cos^2(u))\sin u \ du$$

Then I did u-substitution again and got it to this point:

$$g= \cos u$$ $$-dg=\sin u\ du$$

$$I=-\frac{1}{a}\int(1-g^2)dg$$ $$I=-\frac{1}{a}[g-\frac{g^3}{3}]+C$$ $$I=\frac{1}{a}[\frac{\cos^3(ax)}{3}-\cos(ax)]+C$$

• You mean $d\theta$ where you wrote $dx$? – J. W. Tanner Apr 15 '19 at 21:09
• This looks correct. – Aaron Apr 15 '19 at 21:10
• or $x$ where you wrote $\theta$ ? – J. W. Tanner Apr 15 '19 at 21:12
• @J.W.Tanner I fixed it. – EnlightenedFunky Apr 15 '19 at 21:13
• You, personally, can check indefinite integrals! Strange but true! Differentiate your result and see if it is equal to the original integrand. – GEdgar Apr 15 '19 at 21:47

You can also use linearisation of powers of $$\cos(ax)^n,\sin(ax)^n$$ since it is easy to integrate $$\sin(nax)$$ or $$\cos(nax)$$.
For instance here $$\begin{cases} 4\cos(ax)^3 = 3\cos(ax)+\cos(3ax)\\4\sin(ax)^3 = 3\sin(ax)-\sin(3ax)\end{cases}$$
From which you get $$\int \sin(ax)^3=\frac 14\left(\dfrac{-3\cos(ax)}{a}+\dfrac{\cos(3ax)}{3a}\right)+C=\dfrac{\cos(ax)^3}{3a}-\dfrac{\cos(ax)}{a}+C$$
This technique is very efficient for small powers, because linearising is a time consuming operation since you need to develop $$(\frac{e^{iax}+e^{-iax}}2)^n$$ via binomial formula, and especially if after integration you want the result in the form of powers of $$\cos,\sin$$ you need to factorise it back from its linearised form.