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:

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

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

Is my answer correct?

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

Yes, this is correct. See here for some more general integrals like this.


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.

For higher powers, your technique give a more direct induction formula as shown in the pdf linked by clathratus.


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