# Another Definite Integral Problem

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• Apr 4th 2009, 11:37 AM
fattydq
Another Definite Integral Problem
The definite integral of theta times sqrt(1-cos(2theta) evaluated from 0 to pi/2.

Again with this one, it looks like it'd be an integration by parts problem since there's two unrelated things being multiplied, but the square root seems to make that method almost impossible. Could somebody clue me in on where to begin with this one?
• Apr 4th 2009, 11:42 AM
Jester
Quote:

Originally Posted by fattydq
The definite integral of theta times sqrt(1-cos(2theta) evaluated from 0 to pi/2.

Again with this one, it looks like it'd be an integration by parts problem since there's two unrelated things being multiplied, but the square root seems to make that method almost impossible. Could somebody clue me in on where to begin with this one?

Use a trig. identity $\sqrt{1-\cos(2 \theta)} = \sqrt{1 - (1-2\sin^2 \theta)}$.
• Apr 4th 2009, 11:48 AM
fattydq
Quote:

Originally Posted by danny arrigo
Use a trig. identity $\sqrt{1-\cos(2 \theta)} = \sqrt{1 - (1-2\sin^2 \theta)}$.

Yeah but then I still have a theta out front, and this just seems to complicate the problem...
• Apr 4th 2009, 11:53 AM
Jhevon
Quote:

Originally Posted by fattydq
Yeah but then I still have a theta out front, and this just seems to complicate the problem...

after applying what Danny said, you can use integration by parts. then it's not that complicated
• Apr 4th 2009, 11:56 AM
fattydq
Quote:

Originally Posted by Jhevon
after applying what Danny said, you can use integration by parts. then it's not that complicated

It seems pretty complicated, what would I let be u and what would I let be dv? Because it seems like both terms would be pretty complex to find derivatives/antiderivatives of...
• Apr 4th 2009, 12:02 PM
Jhevon
Quote:

Originally Posted by fattydq
It seems pretty complicated, what would I let be u and what would I let be dv? Because it seems like both terms would be pretty complex to find derivatives/antiderivatives of...

what do you mean both would be complicated? one of them is a polynomial. if you have a polynomial times a trig function, u is the polynomial. unless you have an inverse trig function, then things might change. but that is not the case here
• Apr 4th 2009, 12:05 PM
fattydq
Quote:

Originally Posted by Jhevon
what do you mean both would be complicated. one of them is a polynomial. if you have a polynomial times a trig function, u is the polynomial.

So u would just be theta, and then du would be 1,

and dv would be sqrt(1-(1-2sin^2(theta)) and how the heck am I supposed to find an antiderivative of that!? (Surprised)
• Apr 4th 2009, 12:10 PM
Jhevon
Quote:

Originally Posted by fattydq
So u would just be theta, and then du would be 1,

and dv would be sqrt(1-(1-2sin^2(theta)) and how the heck am I supposed to find an antiderivative of that!? (Surprised)

you were expected to simplify...

$\sqrt{1 - (1 - 2 \sin^2 \theta)} = \sqrt{2 \sin^2 \theta} = \sqrt 2 \sin \theta$

$\sqrt 2$ is a constant, just factor it out and put it in front of the integral sign. i trust you know how to integrate $\sin \theta$
• Apr 4th 2009, 12:15 PM
fattydq
Ahhh, that makes perfect sense, it's always simple algebra things like that that I don't even think of! Thanks a lot man! Just to make sure I'm on the right track, after what you've told me I'll be left with sqrt(2)times the integral of thetasin(theta), using u as theta and dv as sin(theta), correct?
• Apr 4th 2009, 12:17 PM
Jhevon
Quote:

Originally Posted by fattydq
Ahhh, that makes perfect sense, it's always simple algebra things like that that I don't even think of! Thanks a lot man! Just to make sure I'm on the right track, after what you've told me I'll be left with sqrt(2)times the integral of thetasin(theta), using u as theta and dv as sin(theta), correct?

yes
• Apr 4th 2009, 12:19 PM
fattydq
Quote:

Originally Posted by Jhevon
yes

Now since it's a definite integral, at the very last step of the problem when I'm left with sqrt(2) times theta(cos(theta)+sin(theta), do I plug in the values for ALL theta's present or just the sin theta since it's the only part that was still an integral?

I believe I'd plug it in for ALL thetas and my final answer would be 1.4142 etc...can anyone confirm or deny this?

Thanks a lot by the way jhevon
• Apr 4th 2009, 12:30 PM
Jhevon
Quote:

Originally Posted by fattydq
Now since it's a definite integral, at the very last step of the problem when I'm left with sqrt(2) times theta(cos(theta)+sin(theta), do I plug in the values for ALL theta's present or just the sin theta since it's the only part that was still an integral?

I believe I'd plug it in for ALL thetas and my final answer would be 1.4142 etc...can anyone confirm or deny this?

Thanks a lot by the way jhevon

it would be $\sqrt 2 \left( \sin \theta - \theta \cos \theta \right) \Bigg|_0^{\frac {\pi}2}$

your answer should not have any decimals. you should know the exact value of trig functions for those limits. and yes, you plug in the limits (according to the fundamental theorem of calculus) for all $\theta$'s
• Apr 4th 2009, 12:46 PM
Krizalid
It's worth to mention that is actually $\sqrt{2}\left| \sin \theta \right|,$ but for $0\le\theta\le\frac\pi2,$ (i.e. the first quadrant), sine is positive there and the absolute value bars are irrelevant, hence it's $\sqrt{2}\sin \theta .$
• Apr 4th 2009, 12:57 PM
Jhevon
Quote:

Originally Posted by Krizalid
It's worth to mention that is actually $\sqrt{2}\left| \sin \theta \right|,$ but for $0\le\theta\le\frac\pi2,$ (i.e. the first quadrant), sine is positive there and the absolute value bars are irrelevant, hence it's $\sqrt{2}\sin \theta .$

indeed. i was planning on saying that afterward, but forgot :p
• Apr 4th 2009, 02:38 PM
Jester
Sure gald it wasn't $\int_0^{\pi} \theta \,\sqrt{1 - \cos 2 \theta}\, d \theta$ - that would have been mean. (Rofl)