How to Integrate (Cos X)^2?

[sony]

Hi, I just don't get this. I'm to lazy to type inn what I've done, so I just took
a picture of my textbook:
http://home.no.net/erfr1/images/1.jpg
http://home.no.net/erfr1/images/2.jpg
You're supposed to end up with the starting point, right? So you can divide the rest by two?

Whatever I try, I get zero... I got the solution from mathematica, but don't understand a thing of it.


Thanks!

 

[Muzza]

I'm not exactly clear on what it is you have done, but I'm guessing that you tried to integrate cos^2(x) using partial integration, and the equation you got reduced to 0 = 0? I suppose you expected to get back your original integral after a few iterations, so that you could solve for it. I wouldn't say that's what's "supposed" to happen (but it does happen, but not always, as you've demonstrated).

Instead of partial integration, use the identity cos(2x) = 2cos^2(x) - 1 <=> cos^2(x) = cos(2x)/2 + 1/2.

 

[sony]

Yes, I tried partial integration. And thank you for the help!

Cheers

 

[mathwonk]

integration by parts does work of course but only if in the second step, you refrain from undoing the work of the first step. this can be confusing and is actually easier to do by guesswork.

i.e. the derivative of sincos is cos^2 - sin^2. but sin^2 + cos^2 = 1 is also easy to get as a derivative, namely it is the derivative of x. so the derivative of x + sincos is 2cos^2. now you are done after dividing by 2.

of course you notice here that cos^2 - sin^2 = cos(2x) is also coming in as in the trick suggested above, but here you do not have to know that trick.

 

[dextercioby]

Check post number 7 (mine :tongue2: ) from this thread:sine&cosine squared
I think it sould be pretty clear... U have both the primitives/antiderivatives and the definite integrals of the 2 functions wrt to the limits -\pi/2 and +\pi/2.

Daniel.

 

[t!m]

Make sure you read what Muzza said at the end. Just using the identity $$\text{cos}^2(x) = \frac{1}{2}(1+\text{cos}(2x))$$ makes this a very simple integral.

 

[mathwonk]

but the point is not everyone has this identity at their disposal.

 

[TenaliRaman]

Really! I thought it was one of the basic identities in trigonometry usually referred to as the double angle formulae.

-- AI

 

[itchy8me]

hi,

i must integrate sin^2(x) by partial integration.

i I've done this by taking the 2nd partial ingeral and substituting it with the original integral of sin^2(x); and calculate 0 =0. what is it that I'm doing wrong?

what does mathwonk mean by : "integration by parts does work of course but only if in the second step, you refrain from undoing the work of the first step" ?

 

[Geekchick]

why must you use integration by parts? When it's very easy to integrate using the power reducing formula.

 

[itchy8me]

why must you use integration by parts? When it's very easy to integrate using the power reducing formula.

as far as i know, the power reduction formula is acquired by integration by parts. I must integrate by parts, because it can be done, and that is what the excercise says i must do: integrate using partial integration.

 

[itchy8me]

got it : http://www.nevada.edu/~cwebster/Teaching/Notes/Calculus/Integration/intparts.html

i guess mathwork meant that one shouldn't substitute the original integral into the second partial integral...?

 

[itchy8me]

Doh!.. that's exactly the same method i was using, i still get 0=0 thus :(

 

[Count Iblis]

So, it takes more than 4 years for Physics Forums to compute this integral? :uhh:

 

[itchy8me]

So, it takes more than 4 years for Physics Forums to compute this integral? :uhh:

hahahah

i'm wondering how seriously i should take the "integrate by parts" bit, if i do a substitution in the second partial integral with a trigonometric identity, would that be considered cheating?

 

[Geekchick]

you can integrate by parts as long as you use the Pythagorean identity. I don't see how that would be cheating.

 

[itchy8me]

you can integrate by parts as long as you use the Pythagorean identity. I don't see how that would be cheating.

well my logic says that i might just as well then substitute sin^2(x) with an identity in the begining. So my question is now, is this integral solvable with integration by parts alone, or MUST i do a substitution somewhere along the line to beable to solve it?

 

[Geekchick]

Yes, never mind my earlier post I just made the process longer. You can absolutely do it by parts alone. Scratch that I messed up...Give me a minute

 

[Geekchick]

Nope, I cannot see how it can be done without using any identities. As I said though your still using integration by parts.

 

[itchy8me]

Nope, I cannot see how it can be done without using any identities. As I said though your still using integration by parts.

so any hints in how i would do it with integration by parts alone?

 

[Count Iblis]

well my logic says that i might just as well then substitute sin^2(x) with an identity in the begining. So my question is now, is this integral solvable with integration by parts alone, or MUST i do a substitution somewhere along the line to beable to solve it?

Partial integration alone will do. Consider the more general integral of cos^n(x). Let's abbreviate cos(x) by c, sin(x) by s and the integral of cos^n(x) by I_n. We can write:

c^n = c^(n-2) c^2 = c^(n-2)[1-s^2] = c^(n-2) - c^(n-2)s^2

Integrating both sides gives:

I_n = I_{n-2} - Integral of c^(n-2)s^2 dx

We can write:

c^(n-2)s^2 dx = -c^(n-2)sdc

So, we have:

I_n = I_{n-2} + Integral of c^(n-2) s dc (1)


We can write:

c^(n-2) s dc = 1/(n-1)sd[c^(n-1)] =

1/(n-1) {d[sc^(n-1)] - c^(n-1)ds}

The integral can thus be written as:

Integral of c^(n-2) s dc =

1/(n-1) s c^(n-1) - 1/(n-1) Integral of c^(n-1)ds

In the last term we can write ds = cdx, so we have:


Integral of c^(n-2) s dc = 1/(n-1) s c^(n-1) - 1/(n-1) I_{n}

Substituting in Eq. (1) gives:


I_n = I_{n-2} + 1/(n-1) s c^(n-1) - 1/(n-1) I_{n} -------->

n/(n-1) I_n = I_{n-2} + 1/(n-1) s c^(n-1) ---------->

I_{n} = (n-1)/n I_{n-2} + 1/n s c^(n-1)


I_0 is the integral of 1 which is x. So, we have:

I_2 = 1/2 I_0 + 1/2 s c = 1/2 x + 1/2 sin(x)cos(x)

 

[Geekchick]

Count Iblis, You didn't ingrate by parts alone you used the Pythagorean identity in the first part.

 

[Cyosis]

Barring any mistakes from my side you can do it without using any trigonometric identities:

$$\begin{align*} \int \cos^2 x\,dx &= \int 1*\cos^2 x\,dx \\ &= x \x cos^2 x + \int 2x \cos x \sin x\,dx \\ &= x \cos^2 x+(x \sin^2 x- \int \sin^2 x\,dx) \\ &= x \cos^2x + x \sin^2 x - (-\sin x \cos x- \int -\cos^2 x\,dx) \\ &= x (\cos^2x + \sin^2 x) +\sin x \cos x- \int \cos^2 x\,dx \end{align*}$$

It follows that:

$$2\int \cos^2 x\,dx=\x x(cos^2x + \sin^2 x) +\sin x \cos x \Rightarrow$$

$$\begin{align*} \int \cos^2 x\,dx &= \frac{1}{2} (\x x(cos^2x + \sin^2 x) +\sin x \cos x) \\ &= \frac{1}{2} (\x x +\frac{1}{2} \sin {2x}) \\ &= \frac{x}{2}+\frac{1}{4} \sin {2x} \end{align}$$

This is the same result you get by using the double angle identity from the start, as it should be.

 

[itchy8me]

Barring any mistakes from my side you can do it without using any trigonometric identities:

$$\begin{align*} \int \cos^2 x\,dx &= \int 1*\cos^2 x\,dx \\ &= x \x cos^2 x + \int 2x \cos x \sin x\,dx \\ &= x \cos^2 x+(x \sin^2 x- \int \sin^2 x\,dx) \\ &= x \cos^2x + x \sin^2 x - (-\sin x \cos x- \int -\cos^2 x\,dx) \\ &= x (\cos^2x + \sin^2 x) +\sin x \cos x- \int \cos^2 x\,dx \end{align*}$$

It follows that:

$$2\int \cos^2 x\,dx=\x x(cos^2x + \sin^2 x) +\sin x \cos x \Rightarrow$$

$$\begin{align*} \int \cos^2 x\,dx &= \frac{1}{2} (\x x(cos^2x + \sin^2 x) +\sin x \cos x) \\ &= \frac{1}{2} (\x x +\frac{1}{2} \sin {2x}) \\ &= \frac{x}{2}+\frac{1}{4} \sin {2x} \end{align}$$

This is the same result you get by using the double angle identity from the start, as it should be.

brilliant! using the coefficient 1. damn that's a sneaky one. thanks guys and girls ;) ! :-D

 

[ntc2415]

Just use substitution. This makes it all so much easier.

First you set u to be cos x$$\int(cos x)$$² dx

then substitute cos x with u x$$\int(u)$$² dx

Then you take the derivative of both sides of u = cos x to get dx = ?

and dx would be $$\stackrel{}{}du/(cos x)$$