# Math Help - Integrate 3cos^2(x)

1. ## Integrate 3cos^2(x)

Hey guys,

Can you please show me a step by step integration for

Find the solution for the differential equation :

3Cos^2(x) , y= Pi , x = Pi/2

Thank you !

2. Originally Posted by ZaZu
Hey guys,

Can you please show me a step by step integration for

Find the solution for the differential equation :

3Cos^2(x) , y= Pi , x = Pi/2

Thank you !
Note that $\cos (2x) = \cos^2 x - \sin^2 x = 2 \cos^2 x - 1 \Rightarrow \cos^2 x = \frac{\cos (2x) + 1}{2}$.

3. Thanks for the reply, but where did the Cos2(x) come from ?!

4. Originally Posted by ZaZu
Thanks for the reply, but where did the Cos2(x) come from ?!
Are you familiar with the double angle formulae, specifically the fact that $\cos (2x) = \cos^2 x - \sin^2 x ?$ That was my starting point.

5. Yes I do, but I dont understand how using that will affect my answer :S

6. Originally Posted by ZaZu
Yes I do, but I dont understand how using that will affect my answer :S
For goodness sake! I have shown you how to write $\cos^2 x$ in terms of things you should be able to integrate.

Do you know how to integrate $\frac{1}{2} (\cos (2x) + 1) ?$

7. Is it 1/4 Sin2(x) + x ?

8. Originally Posted by ZaZu
Is it 1/4 Sin2(x) + x ?
Only half right. You needed to realise that $\frac{1}{2} (\cos (2x) + 1) = \frac{1}{2} \cos (2x) + \frac{1}{2} \, ....$

9. Oh sorry didnt realize the 1/2 is multiplied by the whole thing in brackets

So its 1/4 Sin2(x) + x/2

10. Originally Posted by ZaZu
Oh sorry didnt realize the 1/2 is multiplied by the whole thing in brackets

So its 1/4 Sin2(x) + x/2
+ C.

And now your job is to find the value of C using the given data.

11. Ohhhh okay got it

So its a rule that I should know, when I have an even power, I should double the angle to get an easier function right ???

12. ummm, sort of. You do have to use double angle formulae if that's what you mean.

Think of this:

$\int cos^4 \theta \ d \theta=\int (cos^2 \theta)^2 \ d \theta$

We already know that (i'll show how Mr F gets his answer since you've never seen it before):

$cos^2 \theta +sin^2 \theta=1$ and $cos^2 \theta - sin^2 \theta= cos 2\theta$.

We require an expression for $cos^2 \theta$.

$cos^2 \theta +sin^2 \theta=1 \Rightarrow \ sin^2 \theta = 1-cos^2 \theta$ from the first identity.

Putting this into the second identity gives:

$2cos^2 \theta -1=cos 2 \theta \Rightarrow \ cos^2 \theta = \frac{cos 2 \theta +1}{2}$.

Going back to our original integral:

$\int cos^4 \theta \ d \theta=\int (cos^2 \theta)^2 \ d \theta = \ \int \left( \frac{cos 2 \theta +1}{2} \right)^2 d \theta$ $=\frac{1}{4} \int (cos 2 \theta +1)^2 d \theta=\frac{1}{4} \int cos^2 2 \theta +2 cos 2 \theta +1 d \theta$.

The big problem now is the $cos^2 2 \theta$.

Looking at the integral you posted and what I have done here, can you finish this integral off?

(Hint: $cos^2 (2 \theta)+sin^2 (2 \theta)=1$ and $cos^2 (2 \theta)- sin^2 (2 \theta)=cos(4 \theta)$).

13. Thanks, Ill get on with it and ill reply when im done