# Thread: How to solve Cos^5 or sin, or similiar (integrals)

1. ## How to solve Cos^5 or sin, or similiar (integrals)

hey guys, I have uploaded the problem below
The problem is \$cos^5x dx
The main part I don't understand is how is the symmetry is done for the exponentials.
right the constants are taken from the pascal triangle. Why is there a part in the formula multiplied with a postive e function, multiplied with a negative
Please don't just tell me to study they book, because I tried that before coming here, and there are no examples on this in ours, they have just given the formula, and thats it, they expect us to know this from highschool. The strange thing here in sweden is that you basically can be admitted to an engineering program without studying complex numbers (math e) I just studied Math D which includes derivatives and basic integrals. They are going to change the admission requirements for engineering programs from autumn this year, so that students with Math E only can get admitted, due to the lack of knowledge... Sucks i know, but what else can I do than coming here.

2. Why not use this resource?
Be sure to click the "show steps".

3. It's easier to use some identities and a u-sub: $\int \cos^5(x) = \int \cos^4(x) \cos(x)$ and that $\cos^2(x) = 1-\sin^2(x) \implies \cos^4(x) = (1-\sin^2(x))^2$

Let $u = \sin(x) \implies \dfrac{du}{dx} = \cos(x) \text{ and } dx = \dfrac{du}{\cos(x)}$ .

$\displaystyle \int \cos^5(x) = \int \cos^4(x) \cos(x) = \int (1-u^2)^2 \cos(x) \dfrac{du}{\cos(x)} = \int (1-u^2)^2 du$ which you can integrate simply enough.

This should work for any odd integer power of $\cos(x)$

4. If you are looking for a complex look at this, then I'll explain your picture for you:

The transition from trig to a complex form is done using Euler's Equations:

$e^{ix} = cos(x) + isin(x)$
$e^{-ix} = cos(-x) + isin(-x) = cos(x) - isin(x)$

Now if you add these two, you get:

$e^{ix}+e^{-ix} = cos(x) + cos(x) + isin(x) - isin(x) = 2cos(x)$

Divide by two to finish it off:

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

As for the red box, which has:

$\int e^{iax} ~dx = \frac{e^{iax}}{ia} + C$

All this is is a REFERENCE to what the integral of e to some power is so that you can separate the sums and apply the formula above.

Everything else you seemed to understand, so if you have any more questions feel free to ask. If you like the reduction formulae presented above though, those are also good ways to solve general powers of cosine and sine.

5. Originally Posted by e^(i*pi)
It's easier to use some identities and a u-sub: $\int \cos^5(x) = \int \cos^4(x) \cos(x)$ and that $\cos^2(x) = 1-\sin^2(x) \implies \cos^4(x) = (1-\sin^2(x))^2$

Let $u = \sin(x) \implies \dfrac{du}{dx} = \cos(x) \text{ and } dx = \dfrac{du}{\cos(x)}$ .

$\displaystyle \int \cos^5(x) = \int \cos^4(x) \cos(x) = \int (1-u^2)^2 \cos(x) \dfrac{du}{\cos(x)} = \int (1-u^2)^2 du$ which you can integrate simply enough.

This should work for any odd integer power of $\cos(x)$

This means that i basically can use this way to solve sin^3x, cos^3x , cos^7x etc, and make a substitution like that everytime? What about even powers solving this kind of question without complex numbers?

Thanks a lot for your explanation guys.. e pi, Aryth and plato

6. All odd powers for sine and cosine can be done with a similar method. Sine to the fifth power has a very similar form (I think it turns out to be $\int -(1 - u^2)^2 ~du$ where $u = cos(x)$).

For even powers all you do is use the following substitutions:

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

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

Since you are reducing to a squared power, you'll eventually come to this:

$\int (cos^2(x))^n ~dx$ where n is even because the initial power was even.

or

$\int (sin^2(x))^n ~dx$ (Again, n is even).

Substitution and solution from here should be clear.