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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?

  • 0
    @Ross: and there's `\quad` and `\qquad` as well for spacing.2011-04-17

2 Answers 2

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

3

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.