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$\int{96\cos^4(16x)} \ dx$
Setting $u=16x$, $du=16dx$, ${\frac{96}{16}}\int{\cos^4(u)} \ du$

Kinda stuck here, I checked Wolfram Alpha but it suggests using some reduction formula that we haven't learned in my class yet.

@Myself: Using what you posted, I end up getting ${\frac{3}{2}}\left(\frac{\sin4u}{4}+\frac{3u}{4}\right)$ Sub back in my $u$ and I've got: $\frac{3\sin(64x)}{8} + \frac{9x}{8}$

I don't think I did it right...

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    The one at the very top was still wrong, as was the one in the title. But I'll fix them.2011-02-10

2 Answers 2

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[note: this is wrong! see below]

Since $\cos 2u = 2\cos^2 u -1$ we have that $ \cos^4 u = \frac{\frac{\cos 4u +1}{2}+1}{2} = \frac{\cos 4u}{4} + \frac{3}{4}$

This should get you started.

[edit] Woops there's a big mistake in here! :-) Let's try again and start from $\cos 2v = 2\cos^2 v - 1$ therefore $ \cos^2 v = \frac{1 + \cos 2v}{2}$ Now squaring on both sides $ \cos^4 v = \frac{ 1 + 2\cos 2v + \cos^2 2v }{4} $ Now use $\cos^2 2v = (\cos 4v +1)/2$: $ \cos^4 v = \frac{ 1 + 2\cos 2v + \frac{1 + \cos 4v}{2}}{4}$ So the correct formula is: $ \cos^4 v = \frac{3}{8} + \frac{1}{2}\cos(2u) + \frac{1}{8}\cos(4u)$ Now all terms can be easily integrated.

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    You are absolutely right to question my logic, sorry about that! I don't know what I was thinking, but I editted my answer. Now it should make more sense.2011-02-10
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Showing a bit more to the end result, we would see the following:

So we are trying to integrate the following expression $~~~\rightarrow ~~~ \dfrac{96}{6} \displaystyle\int \cos^{4} (16x)\ dx$.

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

$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\dfrac{96}{6}\displaystyle\int \cos^{4} (16x)\ dx$

Let: $~u =16x$

$du=16\ dx$

$dx=\dfrac{1}{16}\ du$

Substituting in u and dx we see that we get the following:

$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\dfrac{96}{6}\cdot \dfrac{1}{16}\displaystyle\int \cos^{4} (u)\ du$

Using the reduction formula for cosine to the m power, where $m \in \mathbb{N}$. $\int \cos^{m}(u) dx = \dfrac{1}{m} \cos^{m-1}(u) \sin (u) + \dfrac{m-1}{m} \int \cos^{m-2}(u)\ dx,~ \text{where }~ m = 4,~\text{gives}:$

$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\Rightarrow~\dfrac{96}{6}\cdot \dfrac{1}{16} \Bigg[\dfrac{1}{4} \cos^{3}(u) \sin (u) + \dfrac{3}{4} \displaystyle\int \cos^{2} (u)\ dx \Bigg]$

$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\Rightarrow~\dfrac{1}{4} \cos^{3} (16x) \sin (16x) + \dfrac{3}{4} \displaystyle\int \cos^{2} (u)\ dx $

Now we can use the the trigonometric identity for $\cos^{2}(u)$ and re-write it as the folllowing: $\cos^{2}(u)=\dfrac{1}{2}+\dfrac{1}{2}\cos (2u)$ . Now with this, let's replace the integrand with this identity and $u$ substitution as so,

$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\Rightarrow~\dfrac{1}{4} \cos^{3} (16x) \sin (16x) + \dfrac{3}{4}\displaystyle\int \dfrac{1}{2}+\dfrac{1}{2}\cos (2u)\ du$

Which now we can integrate each separately as they are being added as a sum and also pull out any constants from the integrand as the following:

$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\Rightarrow~\dfrac{1}{4} \cos^{3} (16x) \sin (16x) + \dfrac{3}{4} \Bigg[ \dfrac{1}{2} \displaystyle\int \! \ du + \dfrac{1}{2} \int \cos (2u)\ du \Bigg]$

Now making another substitution, we see the following:

Let: $~w =2u$

$dw=2\ du$

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

Making the substitutions in for w and du we see that we get the following:

$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\Rightarrow~\dfrac{1}{4} \cos^{3} (16x) \sin (16x) + \dfrac{3}{8} \displaystyle\int \ du + \dfrac{1}{2}\cdot \dfrac{1}{2}\cdot \dfrac{3}{4} \int \cos (w)\ dw$

$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\Rightarrow~\dfrac{1}{4} \cos^{3} (16x) \sin (16x) + \dfrac{3}{8} u + \dfrac{3}{16} \sin (w) + K$

Plugging back in for what $w$ is gives:

$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\Rightarrow~\dfrac{1}{4} \cos^{3} (16x) \sin (16x) + \dfrac{3}{8} u + \dfrac{3}{16} \sin (2u) + K$

$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\Rightarrow~\dfrac{1}{4} \cos^{3} (16x) \sin (16x) + \dfrac{3}{8} u + \dfrac{3}{16} \cdot 2\sin (u)\cos (u) + K$

$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\Rightarrow~\dfrac{1}{4} \cos^{3} (16x) \sin (16x) + \dfrac{3}{8} u + \dfrac{3}{8} \sin (u)\cos (u) + K$

Plugging back in for what $u$ is gives:

$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\Rightarrow~\dfrac{1}{4} \cos^{3} (16x) \sin (16x) + \dfrac{3}{8} \cdot (16x) + \dfrac{3}{8} \sin (16x) \cos (16x) + K$

$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\Rightarrow~\dfrac{1}{4} \cos^{3} (16x) \sin (16x) + \dfrac{3}{8} \cdot (16x) + \dfrac{3}{8}\cdot \dfrac{1}{2} \sin (32x) + K$

Which is, $~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\therefore~\dfrac{1}{4} \cos^{3} (16x) \sin (16x) + 6x + \dfrac{3}{16} \sin (32x) + K.~~~~~~~~~~~~~~~~~~~~~~~~\blacksquare$

Which can be cleaned up a bit further to this:

$\dfrac{1}{4}\Bigg[\cos^{3}(16x) \sin (16x) + \dfrac{3}{4}\sin (32x) + 24x\Bigg] + K.$

NOTE: This expression can reduced further using more identities, but not necessary. I will leave it as this stage. Just wanted to point that out in case you see or get a different solution from this here.

Okay, I hope that this has helped out. Let me know if there is any step covered that did not make much sense for doing so.

Thanks.

Good Luck.