Integrate 3cos^2(x)

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• Apr 14th 2009, 12:47 AM
ZaZu
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 !
• Apr 14th 2009, 01:25 AM
mr fantastic
Quote:

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 $\displaystyle \cos (2x) = \cos^2 x - \sin^2 x = 2 \cos^2 x - 1 \Rightarrow \cos^2 x = \frac{\cos (2x) + 1}{2}$.
• Apr 14th 2009, 01:42 AM
ZaZu
Thanks for the reply, but where did the Cos2(x) come from ?!
• Apr 14th 2009, 02:04 AM
mr fantastic
Quote:

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 $\displaystyle \cos (2x) = \cos^2 x - \sin^2 x ?$ That was my starting point.
• Apr 14th 2009, 02:16 AM
ZaZu
Yes I do, but I dont understand how using that will affect my answer :S
• Apr 14th 2009, 02:24 AM
mr fantastic
Quote:

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 $\displaystyle \cos^2 x$ in terms of things you should be able to integrate.

Do you know how to integrate $\displaystyle \frac{1}{2} (\cos (2x) + 1) ?$
• Apr 14th 2009, 02:37 AM
ZaZu
Is it 1/4 Sin2(x) + x ?
• Apr 14th 2009, 02:41 AM
mr fantastic
Quote:

Originally Posted by ZaZu
Is it 1/4 Sin2(x) + x ?

Only half right. You needed to realise that $\displaystyle \frac{1}{2} (\cos (2x) + 1) = \frac{1}{2} \cos (2x) + \frac{1}{2} \, ....$
• Apr 14th 2009, 02:44 AM
ZaZu
Oh sorry didnt realize the 1/2 is multiplied by the whole thing in brackets

So its 1/4 Sin2(x) + x/2
• Apr 14th 2009, 02:47 AM
mr fantastic
Quote:

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.
• Apr 14th 2009, 02:49 AM
ZaZu
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 ???
• Apr 14th 2009, 03:35 AM
Showcase_22
ummm, sort of. You do have to use double angle formulae if that's what you mean.

Think of this:

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

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

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

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

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

Putting this into the second identity gives:

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

Going back to our original integral:

$\displaystyle \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$$\displaystyle =\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 $\displaystyle cos^2 2 \theta$.

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

(Hint: $\displaystyle cos^2 (2 \theta)+sin^2 (2 \theta)=1$ and $\displaystyle cos^2 (2 \theta)- sin^2 (2 \theta)=cos(4 \theta)$).
• Apr 14th 2009, 03:41 AM
ZaZu
Thanks, Ill get on with it and ill reply when im done :)