Another integration by substitution problem

H

hank

Junior Member
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Sep 13, 2006
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209
Ok, here's the problem:

\(\displaystyle \int\limits_{\frac{\pi }{3}}^{\frac{\pi }{2}} {\sin \theta \sqrt {1 - 4\cos ^2 \theta } d\theta }\)

The book tells me to let
\(\displaystyle u = 2\cos \theta\)

Ok.
So I do that, and I get:

\(\displaystyle \frac{1}{2}du=\sin \theta d\theta\)

Which leads me to this after substitution:

\(\displaystyle \frac{1}{2}\int_0^1 {\sqrt {1 - (2u)^2 } }\)

Which is where I get stuck. I'm not sure where to proceed from here.
If I plug in for u, I end up with an undefined square root.

I'm thinking that the problem looks like some sort of an inverse trig function, but I can't place it.
 
Where'd you go astray?. Just make the substitution they suggested.

\(\displaystyle \L\\\int_{\frac{\pi}{3}}^{\frac{\pi}{2}}sin{\theta}\sqrt{1-4cos^{2}{\theta}}d{\theta}\)

\(\displaystyle \L\\u=cos{\theta}; \;\ -du=sin{\theta}d{\theta}\)

\(\displaystyle \L\\-\int_{\frac{1}{2}}^{0}\sqrt{1-4u^{2}}du\)

\(\displaystyle \L\\\int_{0}^{\frac{1}{2}}\sqrt{1-4u^{2}}du\)
 
Sorry, it was supposed to be u = 2 cos theta.

So I end up with:
\(\displaystyle - \frac{1}{2}\int\limits_0^1 {\sqrt {1 - 4u^2 } }\)


What happens is that when you plug 1 in for u^2, you end up with the sqrt of -3, which is undefined, right?
 
That substitution works OK.

After making it, I get:

\(\displaystyle \L\\\frac{1}{2}\int_{0}^{1}\sqrt{1-u^{2}}du\)


You shouldn't have that 4 in there. It is done away with for the \(\displaystyle u^{2}\)

Because \(\displaystyle (2cos{\theta})^{2}=4cos^{2}{\theta}=u^{2}\)
 
No more minus sign out front?

Do I want to sub in a inverse trig function somehow?
 
No, with the minus sign you have:

\(\displaystyle 2cos(\frac{\pi}{3})=1\)

\(\displaystyle 2cos(\frac{\pi}{2})=0\)

You get:

\(\displaystyle \L\\-\frac{1}{2}\int_{1}^{0}\sqrt{1-u^{2}}du\)

Note, the limits are backwards. The largest one is on the bottom.

You can reverse those and eliminate the minus sign.
 
Ok, where did the 1/2 go? Also, do I want to sub in a trig derivative somehow?

If I set the square root to ^-1 and sub in inverse sin for it, does that then become sin?
 
Nothing happened to the 1/2, I just forgot it.

You can use trig sub if you like. \(\displaystyle u=sin{\theta}; \;\ du=cos{\theta}d{\theta}\)

Don't forget to change your limits of integration.
 
Ok, the answer I got was 1/2, but according to the book the answer is pi/8?
 
From the trig substitution, you should have the integral:

\(\displaystyle \L\\\frac{1}{2}\int_{0}^{\frac{\pi}{2}}cos^{2}{\theta}d{\theta}\)

\(\displaystyle \L\\\frac{1}{2}\int_{0}^{\frac{\pi}{2}}(\frac{1+cos2{\theta}}{2})d{\theta}\)

\(\displaystyle \L\\\frac{1}{4}\int_{0}^{\frac{\pi}{2}}(1+cos2{\theta})d{\theta}\)

You should be able to finish now, huh?. Keep trying. You'll get it.
 
Ya, I'm sorry but I don't see how you got from a to b on that one. How did the 1 become a pi/2 on the integral. I'm just totally lost on this one, unfortunately.
 
Of course, I changed my limits of integration.

\(\displaystyle \L\\\frac{1}{2}\int_{0}^{1}\sqrt{1-u^{2}}du\)

Let \(\displaystyle u=sin{\theta}; \;\ du=cos{\theta}d{\theta}\)

This changes the limits of integration. You always have to do that when you change variables, such as in u or trig substitution.

\(\displaystyle {\theta}=sin^{-1}(0)=0; \;\ {\theta}=sin^{-1}(1)=\frac{\pi}{2}\)

These are our new limits.

\(\displaystyle \L\\\frac{1}{2}\int_{0}^{\frac{\pi}{2}}\sqrt{1-sin^{2}{\theta}}cos{\theta}d{\theta}\)

\(\displaystyle \L\\\frac{1}{2}\int_{0}^{\frac{\pi}{2}}\sqrt{cos^{2}{\theta}}cos{\theta}d{\theta}\)

\(\displaystyle \L\\\frac{1}{2}\int_{0}^{\frac{\pi}{2}}cos^{2}{\theta}d{\theta}\)

\(\displaystyle \L\\\frac{1}{2}\int_{0}^{\frac{\pi}{2}}(\frac{1+cos2{\theta}}{2})d{\theta}\)

\(\displaystyle \L\\\frac{1}{4}\int_{0}^{\frac{\pi}{2}}(1+cos2{\theta})d{\theta}\)

Now, surely you can finish.
 
How does this:

\(\displaystyle {\theta}=sin^{-1}(0)=0; \;\ {\theta}=sin^{-1}(1)=\frac{\pi}{2}\)

Factor in? I thought we already changed it which is why we went from pi/3 and pi/2 to 0 and 1. And why use inverse trig functions?

Also, we already defined u as 2cos theta. Why redefine it yet again, and why sin? why not redefine it as anything else?

And, how/why do you go from:

\(\displaystyle \L\\\frac{1}{2}\int_{0}^{\frac{\pi}{2}}cos^{2}{\theta}d{\theta}\)

to:

\(\displaystyle \L\\\frac{1}{2}\int_{0}^{\frac{\pi}{2}}(\frac{1+cos2{\theta}}{2})d{\theta}\)
 
Look up trig substitution in your calc book and/or see your instructor.

Also, you should know the identity \(\displaystyle cos^{2}{\theta}=\frac{1+cos2{\theta}}{2}\)
 
I will do that.

I actually checked my pages of identities, but I'm missing something somewhere.

Guess I just have a mental block on this one.

Thanks very much for trying though!
 
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