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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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Hint: $\cos^2(\theta) = \frac{1 + \cos(2 \theta)}{2}$. –  JavaMan Feb 9 '11 at 19:46
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You should not have $x$ be the variable of the function, and have a different letter in the "$du"$ part. That's just inviting trouble. Read literally, your integral would be equal to $16\cos^4(16x)u + C$. –  Arturo Magidin Feb 9 '11 at 20:56
    
@Arturo: True, I typed it wrong, let me fix it up. How about now? –  Finzz Feb 10 '11 at 1:34
    
The one at the very top was still wrong, as was the one in the title. But I'll fix them. –  Arturo Magidin Feb 10 '11 at 4:38
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2 Answers 2

up vote 4 down vote accepted

[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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Thanks, if it's not too much can you just show how to work out the identity to account for $cos^4(u)$? I see that you would have cos4u instead on the left, and then would you just end up squaring the right side? –  Finzz Feb 10 '11 at 2:03
    
I guess what I'm trying to say is that I see that you would have cos4u instead on the left, and then would you just end up squaring the right side? I see how what you wrote below it works out though. Is the only way to just write it out the long way you did? Because then I see how u added 1 to the left and divided by 2. Then u would be done if you were only working with cos^2(u), but what's the reasoning behind just adding 1 again and dividing by 2 again since u have cos4u and not cos2u. –  Finzz Feb 10 '11 at 2:09
    
Screwed up that second comment, I meant to erase the first sentence. It resulted from the 5 min rule on comment edits, and now I have to write this comment because of the 5 minute rule again. <sigh> –  Finzz Feb 10 '11 at 2:18
    
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. –  Myself Feb 10 '11 at 12:04
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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.

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+1 for the general recurrence. –  Américo Tavares Apr 22 '11 at 10:52
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