Integrate 3cos^2(x): Step by Step Solution

[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 :

3Cos²(x) , y= Pi , x = Pi/2



Thank you !

 

[danago]

To integrate (cos x)^2, can you think of a trig. identity than can be used to change it to a form that should be easy to integrate?

 

[ZaZu]

Ummm Cos2(x) ?? =\

Sin^2(x) ?

To be honest I do not know, that's why I wanted help.

How would I know what Trig function would I need to integrate that ?!

Please help ! :(

 

[ZaZu]

Someone please help me, I have an exam tomorrow and there are several questions with the same style !

 

[danago]

Ok, perhaps start from cos(2x).

Do you know the "double angle formula"? How can you write cos(2x) in terms of just cos(x)? Try to figure that out, and it should be clear what to do next after that.

 

[User Name]

danago, he's asking for cosine squared of x, not cosine of 2x.

ZaZu, have you learned integration by parts?
That's really the only way I can see you integrating this function.

Set u = 3cos^2(x) and dv = dx.

Solve from there.

 

[Hootenanny]

danago, he's asking for cosine squared of x, not cosine of 2x.

ZaZu, have you learned integration by parts?
That's really the only way I can see you integrating this function.

Set u = 3cos^2(x) and dv = dx.

Solve from there.

User Name, it is far more straight forward to transform cos²x into cos(2x) and integrate from there rather than integrating cos²x using integration by parts, which is messy.

 

[ZaZu]

Integration by parts will prompt a more complicated answer :S I tried it and its too messy

I want to know how do we convert Cos^2x into Cos(2x) .. what's the relationship between that ?? Cos2x is a double angle, and cos^2(x) is Cosine Squared ... Arent they both different ?

Please clarify that to me !

 

[danago]

Cos(2x) = Cos²(x)-Sin²(x) = 2Cos²(x)-1

Can you see how to use that?

 

[User Name]

User Name, it is far more straight forward to transform cos²x into cos(2x) and integrate from there rather than integrating cos²x using integration by parts, which is messy.

Ah, my mind completely skipped over that trigonometric property, and simply thought that danago had misread ZaZu's original problem.

My apologies.

Anyway, danago is pushing you towards the correct answer, ZaZu.

 

[ZaZu]

This is what I am getting .. I am using another method ..

I am failing to understand how can I convert Cos^2(x) into Cos2(x) http://img95.imageshack.us/img95/9572/image355.th.jpg

Oh wait, I just realized I solved for Cos^3(x) ... omg *faceplam*

 

[danago]

This is what I am getting .. I am using another method ..

I am failing to understand how can I convert Cos^2(x) into Cos2(x) http://img95.imageshack.us/img95/9572/image355.th.jpg

How did you get from your first to second line?

Cos²(x)=Cos(x)Cos²(x)?

From the trig identity i posted, you can change to:

Cos²(x)= [Cos(2x)+1]/2

Which is much easier to integrate.

 

[ZaZu]

Yeah I mentioned it up there, I just realized I solved for Cos^3(x) xD

The thing is, I don't understand WHY would I convert it into cos2x even if I used the trig function of Cos2(x) = 2Cos^2(x) - 1 ..

 

[danago]

Simply because it is much easier to integrate.

3Cos²(x)= 3[Cos(2x)+1]/2 = (3/2) [Cos(2x) + 1]

Now you can integrate easily using the fact that \int Cos(ax) dx = (1/a) Sin(ax)+C

 

[ZaZu]

So you're saying that its a rule I should memorize ??

 

[danago]

The trig identity? You should learn (either memorize or learn to derive) the common trig identities because they can often be used to make an integration much easier to perform by allowing you to change the integrand to something "nicer". Trig identities are very useful to know, not just for intergration, but for math in general.

 

[ZaZu]

I do know some of the trig identities, but I do not understand how using them here can help me.

But I came up with this rule : If the index is an EVEN number, I double the angle.
If the index is an ODD number, I use the trig identities to substitute instead.

Cos^2(x) = Cos(2x)

Cos^3(x) = Cos(x) x Cos^2(x)
.....= Cos(x) x (1 - Sin^2(x) ) ... etc

 

[Cyosis]

I don't want to sound like an ***, but reading carefully is a very important part of mathematics and science. Multiple people have given you the answer yet you seem to just ignore it and come up with a rule of yourself that is plainly wrong. Coming up with rules yourself is very good, however do try to prove them so you know they are correct.

Cos(2x) = Cos²(x)-Sin²(x) = 2Cos²(x)-1

Can you see how to use that?

Could you solve this equation for \cos^2(x)?

 

[ZaZu]

Cyosis I am having a difficulty understanding what they are telling me, I do not know why, I thinks its nervousness prior to exams. I am panicking ..

Are you asking me to solve Cos^2(x) as in integrate Cos^2(x) ?

By the way, I am sorry everyone, my brain is just not tolerating a thing anymore .. :(

 

[danago]

That first line is not correct. Cos²(x) is not the same as cos(2x).

with the trig identity, what i am saying is that cos²(x) is equivalent to [cos(2x)+1]/2 for all values of x -- they are practically the exact same things just written differently (thats what an identity is). Hence integrating the second form will effectively give you the exact same result as integrating the first form, only difference is that the original form is a lot harder to do.

 

[Cyosis]

Panicking certainly won't help, here goes.
Given the trigonometric identity \cos(2x)=2\cos^2(x)-1. Could you rewrite this equation to \cos^2(x)=...[/itex]?<br /> <br /> Edit: Seeing as you have an exam tomorrow I would really learn the following trig identities if I were you, any calc+ class will expect you to know them.<br /> <br /> &lt;br /&gt; \begin{align*}&lt;br /&gt; &amp;amp; \sin^2 x+\cos^2 x =1&lt;br /&gt; \\&lt;br /&gt; &amp;amp; \cos 2x=2\cos^2 x-1&lt;br /&gt; \\&lt;br /&gt; &amp;amp; \cos2x = 1-2 \sin^2 x&lt;br /&gt; \\&lt;br /&gt; &amp;amp; \sin 2x =2 \sin x \cos x&lt;br /&gt; \end{align}&lt;br /&gt;

 

[ZaZu]

Oh so they are EQUAL ??

oh .. Ohhhh I got it !

Omg thank you so much Danago, and thank you all who replied !

Thanks a lot !

Btw, anyone got a way to avoid being nervous from the exam ? Its really a pain in the head .. I don't want it but I can't help not panicking =\

[EDIT]
@Cyosis

If Cos2x = 2Cos^2(x) -1 ..
Then cos^2(x) = [Cos2(x) +1 ] . 1/2

Which gives 1/2[Cos2(x) +1] ..

Is that correct ??--EDIT

OMG I JUST GOT IT ... We are REARRANGINGGGG ! I mean we are just re-arranging the whole identity !
omg *slaps face*

 

[Cyosis]

Yes that is correct so now you can replace the integrand of cos^2 x with 1/2[Cos2(x) +1] which is easy to integrate!

 

[danago]

Yea they are equal, for all values of x So it all makes sense now?

As for the exam nerves, i think the best way to reduce them is to do as many practice exams as you possibly can. The more you do, the more comfortable you will get with them.

 

[ZaZu]

Oh my God .. !

The whole process was just to re-arrange the identity .. how can I miss that :"(

ahhh ! Thanks guys ! THANK YOU SO MUCH

@ Danago : The thing is I still did not revise the past lessons, so I feel nervous by thinking * Oh I still did not finish this and I still have other lessons ! *
Which does not help at all :(

Im having green tea, which is said to reduce stress ...

Damn those neurons for sending stress signals :@

 

[Cyosis]

To get some more feeling for these type of integrals perhaps it's wise to try \int \sin^2 x\,dx

 

[ZaZu]

To get some more feeling for these type of integrals perhaps it's wise to try \int \sin^2 x\,dx

Yes I will solve that, I am sure I will get the correct answer ..

Regarding your previous post, I already have a list infront of me, which by some unknown way I did not use at all to figure out my problem in the first place..

http://img89.imageshack.us/img89/1392/image356.th.jpg

Thank you so much all of you

Thank you again n again !

 

[ZaZu]

Ok I solved Sin^2(x)

My answer is x - 1/4[Sin2(x)]

However, I think the x is missing the something ..

Did I do something wrong ?

[edit] Got it, the 1/2 is multiplied to the whole side not just Cos2x .. :)

Is my answer correct now ?

x/2 - 1/4 [Sin2(x)]

 

[Cyosis]

Correct.

 

[ZaZu]

Thank you :)