# The integral $\int \cos^3(2x) \, \mathrm dx$

Given the following problem:

integrate $\cos^3(2x)$

I was given the solution

\begin{align*} \int\cos^3(2x)\, \mathrm dx &= \int\cos^2(2x) \cos 2x\, \mathrm dx = \int(1-\sin^2 2x)\cos 2x\, \mathrm dx\\ &= \frac12\int (1-u^2)\, \mathrm du = \cdots \end{align*}

but the problem is I am stuck on the 1/2. Where did it come from?

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$\sin 2x = u$. Therefore, $\cos2x dx = \frac{1}{2}du$. That's where the 1/2 comes from. –  svenkatr Mar 8 '11 at 6:03
+1 for thinking about the problem and showing where you have an issue. I wish I could +1 the edit from Arturo Magidin that makes it so much easier to read. –  Ross Millikan Mar 8 '11 at 6:24
@Arturo: does \, give a little space in $\LaTeX$? I hadn't seen that before and have wanted it. –  Ross Millikan Apr 17 '11 at 15:45
@Ross: Yes: \, gives a "thin space", \; gives a medium space, \> gives a "thick space" (all smaller than \ ; and \! gives a "negative thin space", which is sometimes useful. I usually use \, before the differentials in integrals, and I believe that's the standard... –  Arturo Magidin Apr 17 '11 at 15:47
@Ross: and there's \quad and \qquad as well for spacing. –  Ｊ. Ｍ. Apr 17 '11 at 16:03

When the substitution is made, $u=\sin 2x$, so $du=2\cos 2x\;dx$, or $\frac{1}{2}du=\cos 2x\;dx$.

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So we are trying to integrate the following expression $~~~\rightarrow ~~~ \displaystyle\int \cos^{3} (2x)\ dx$.

To do the this, we will need to make an appropriate substitution inside of the integrand. Doing this leads us to the following:

$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\displaystyle\int \cos^{3} (2x)\ dx$

Let: $~u =2x$

$du=2\ dx$

$dx=\dfrac{1}{2}\ du$

$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\dfrac{1}{2}\displaystyle\int \cos^{3} (u)\ du$

Using the reduction formula, $$\int \cos^{m}(u) du = \dfrac{1}{m} \cos^{m-1}(u) \sin (u) + \dfrac{m-1}{m} \int \cos^{m-2}(u)\ du,~ \text{where }~ m = 3,~\text{gives}:$$

$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\Rightarrow~\dfrac{1}{2}\Bigg[\dfrac{1}{3} \cos^{2}(u) \sin (u) + \dfrac{2}{3} \displaystyle\int \cos (u)\ du \Bigg]$

$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~=~\dfrac{1}{6} \cos^{2} (2x) \sin (2x) + \dfrac{1}{3} \sin (2x) + C~~~~~~~~~~~~~~~~~~~~~~~~~~~~\blacksquare$

Which can be simplified further to this:

$$\dfrac{1}{24}\Bigg(9 \sin (2x) + \sin (6x)\Bigg) + C.$$

Okay, I hope that this has helped out and now you see where the $\dfrac{1}{2}$ came from. Let me know if there is any step covered that did not make much sense for doing so.

Thanks.

Good Luck.

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