Integrate 3cos^2(x)

**ZaZu** · April 14th 2009, 12:47 AM

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 !

**mr fantastic** · April 14th 2009, 01:25 AM

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

**ZaZu** · April 14th 2009, 01:42 AM

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

**mr fantastic** · April 14th 2009, 02:04 AM

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.

**ZaZu** · April 14th 2009, 02:16 AM

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

**mr fantastic** · April 14th 2009, 02:24 AM

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

**ZaZu** · April 14th 2009, 02:37 AM

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

**mr fantastic** · April 14th 2009, 02:41 AM

Originally Posted by ZaZu

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

Only half right. You needed to realise that $\frac{1}{2} (\cos (2x) + 1) = \frac{1}{2} \cos (2x) + \frac{1}{2} \, ....$

**ZaZu** · April 14th 2009, 02:44 AM

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

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

**mr fantastic** · April 14th 2009, 02:47 AM

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.

**ZaZu** · April 14th 2009, 02:49 AM

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 ???

**Showcase_22** · April 14th 2009, 03:35 AM

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 $cos^2 (2 \theta)- sin^2 (2 \theta)=cos(4 \theta)$ ).

**ZaZu** · April 14th 2009, 03:41 AM

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

Thread: Integrate 3cos^2(x)

LinkBack

Thread Tools

Display

Integrate 3cos^2(x)

Similar Math Help Forum Discussions

[SOLVED] Linear Approximation of 3cos^2(3) to estimate the value of |cos(3)

Find all the zeros of this function: f(x) = 1 - 3cos(2x) in this interval?

Derivative of 3cos^5x.

graphing [math]y=3cos(x+3pi)-2[/math]

Classify & solve 2y' = tan(x)/(y-2y^3cos(x))

Search Tags