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$$\int \cos^4 (2t)dt$$

$u = 2t$ $$\frac {1}{2}\int \cos^4 u du$$ $$\frac {1}{2}\int \frac{(1+ \cos^2u)^2}{2} du$$

$$\frac {1}{8}\int 1 - \cos^2 2u du$$

$$ \frac{1}{8} \int \sin^2 u du$$

From here I am not sure what to do. Nothing seems to simplify the problem, only complicate it.

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    [Well ...](http://www.wolframalpha.com/input/?i=integral+cos%5E4+2t+dt)2012-06-01
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    You can expand $\left(\cos^2(x)\right)^2$ into an expression in terms of $\left(\cos(2x)\right)^2$, and then into an sum of terms of the form $a\cos nx$ for various $n≤4$. Then you can integrate each of these separately.2012-06-01
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    I strongly suggest you go in person to a tutor. You have asked nearly $10$ questions today on more or less the same sort of integral. From your comment to answers, you also don't seem to understand the overall principle and you are only interested in a quick fix solution.2012-06-01
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    I might find time on Monday for a tutor but I have found tutors to be mostly useless and incredibly expensive.2012-06-01
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    @Jordan: I can give you my email so we'll work on the problems together. Let me know if you need it or you're doing well on your own.2012-06-03
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    @Gigili Thanks but I wouldn't want to waste that much of your time, I think I can learn enough from just asking questions here and maybe a tutor on monday.2012-06-03

2 Answers 2

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$$\cos^2 x=\frac{1}{2}(\cos(2x)+1)$$

Thus

$$ f(x)=\cos^4 x=\left(\frac{1}{2}(\cos(2x)+1)\right)^2=\frac{1}{4}\left(\cos^2(2x)+2\cos(2x)+1\right)=\frac{1}{8}(4\cos(2x)+\cos(4x)+3) $$

And we may easily integrate this.


An alternate (but less efficient) solution is to do as follows:

$$\cos^4 x=\left(\frac{e^{ix}+e^{-ix}}{2}\right)^4=\frac{1}{16}(e^{ix}+e^{-ix})^4$$

Expanding, we have

$$\cos^4 x=\frac{1}{16}(6+4e^{-2ix}+4e^{2ix}+e^{4ix}+e^{-4ix})$$

Which again may be integrated.

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    Shouldn't the third equation be $\frac{1}{4}$ instead of $1$?2012-06-01
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    @Gigili Indeed, forgot the brackets!2012-06-01
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    I can't easily integrate that.2012-06-01
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    @Jordan: Why not? Think of integrating as anti-derivative. You have some terms with $\cos$ only.2012-06-01
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    I have 2cos2x and cos^2 2x I have no idea what choices to make to integrate these smaller terms, and how do I know for sure I am correct up to this point? I could use a double angle which has 3 identities and I do not know which one to use.2012-06-01
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    @Jordan There are no more exponents! Just make two simple substitutions of $u=2x \implies du=2$ and $u=4x \implies du=4$. The rest of the stuff are just constant coefficients and can be ignored,2012-06-01
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    @Argon I am not sure if I am doing this correctly but I am trying to solve 2cos2x so I made it into $2\int cos^ 2 x - sin^2 x$ and then I broke that up further and I am trying to solve both terms seperately but this is incredibly difficult.2012-06-01
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    @Jordan I see you are trying to use a double angle, but that is just confusing things (you end up again with $\cos^2 x$ too). Try this: $2x=u \implies 2 dx = du$. Making the substitution: $$2\frac{1}{2}\int \cos u \, du=\sin u + C = \sin (2x)+C$$2012-06-01
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    This problem is too hard I just can't do it, I have to go back to a more simple problem. There is just too much to go wrong in this problem.2012-06-01
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Well, $$\int \sin^2 u du = \int 1-\cos^2 u du = \int 1-\frac{1}{2}\left(1+\cos 2u \right) du$$ using formula for the double angle found for example here: http://www.sosmath.com/trig/douangl/douangl.html

Now, you can probably continue, using for example substitution $$2u = t $$.

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    Some parenthesis will help.2012-06-01
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    I am working with sin not cos.2012-06-02