0
$\begingroup$

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

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

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

$ \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.

  • 0
    @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

2

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

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.

  • 0
    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
0

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 $.

  • 0
    I am working with sin not cos.2012-06-02