integrate cos^(1/2)sin^3x dx from 0 to pi/2

- anonymous

integrate cos^(1/2)sin^3x dx from 0 to pi/2

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- zepdrix

\[\Large \int\limits_0^{\pi/2}\sqrt{\cos x}\;\sin^3x\;dx\]So this is what we've got to solve? Hmm

- anonymous

Yes, so far I've pulled out one of the sinx then changed the \[\sin ^{2}x \to (1-\cos ^{2}x)\]

- anonymous

then I guess you do a u sub and u=cosx and du would be -sinx

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- anonymous

but I'm not sure where to go from there

- zepdrix

-du=sinx?
looks like you're on the right track

- zepdrix

\[\Large \int\limits_0^{\pi/2}\cos^{1/2}x\left(1-\cos^2x\right)(\sin x\;dx)\]
\[\Large u=\cos x \qquad\to\qquad -du=\sin x\;dx\]

- zepdrix

If I wrote it like this, does it become a tad clearer?\[\Large -du=(\sin x\;dx)\]

- zepdrix

Maybe you can match it up with the problem that way hehe

- anonymous

ahh, I see what you're saying

- anonymous

then the negative on that would just come out from like a constant

- zepdrix

I would probably distribute the negative to each term in the (1-u^2) brackets.
But yah, you can pull it out of the integral if you want.

- anonymous

and the dx's would cancel. So you'd be left with \[-\int\limits_{0}^{\pi/2} u ^{1/2}(1-u ^{2}) du\]

- zepdrix

Looks good, let's just not forget that these limits of integration don't apply to our u variable.\[\Large -\int\limits\limits_{\color{red}{x=0}}^{\color{red}{\pi/2}} u ^{1/2}(1-u ^{2}) du\]

- zepdrix

writing x= is a good way to remind yourself of that.

- anonymous

oh, okay. Yeah, I usually forget that :/

- zepdrix

So what's the next step? \c:/

- anonymous

I would probably distribute the \[u ^{1/2}\] to the parenthesis then solve the equation like seperate integrals

- zepdrix

sounds good :O

- anonymous

so now I have \[-(\frac{ 2 }{ 3 })u ^{3/2}-(\frac{ 1 }{ 3 })u ^{3}\]

- anonymous

or did I forget something?

- zepdrix

Ah sorry I disappeared there :3
I would have probably distributed this negative into the brackets as well as distributing the u^1/2.
I think you're less likely to make a mistake that way :o
\[\Large -\int\limits\limits\limits_{\color{red}{x=0}}^{\color{red}{\pi/2}} u ^{1/2}(1-u ^{2}) du \qquad\to\qquad \Large \int\limits\limits_{\color{red}{x=0}}^{\color{red}{\pi/2}} u ^{5/2}-u^{1/2}\; du\]

- zepdrix

\[\Large u^{1/2}u^{2} \quad=\quad u^{5/2}\]

- anonymous

apparently I can do calculus but can't add :/ lol

- zepdrix

hehe

- anonymous

but did I have the rest correct? then I'd sub back for the u?

- zepdrix

AHHHH stupid openstudy freezing >:(

- zepdrix

One other small mistake, when you distribute that negative, you should have gotten a +on the second term.\[\Large -\frac{2}{3}u^{3/2}\color{red}{+}\frac{2}{7}u^{7/2}\]

- anonymous

yeah :/

- anonymous

I can't complain too much though. Don't know what I'd do without it

- zepdrix

From there, ya let's undo our substitution, then we can evaluate the result at our limits.

- anonymous

ah, okay

- anonymous

\[\left[ -\frac{ 2 }{ 3 } \right]cosx ^{3/4}+\frac{ 2 }{ 7 }\cos ^{7/2}\] evaluated at 1 to 0?

- anonymous

\[\left[ -\frac{ 2 }{ 3 } \right]cosx ^{3/4}+\frac{ 2 }{ 7 }\cos ^{7/2}\] evaluated at 1 to 0?

- zepdrix

The first one is to the 3/2 power right? :o

- anonymous

yeah, typo

- zepdrix

Evaluated from 0 to pi/2.
\[\Large \frac{2}{7}\cos^{7/2}x-\frac{2}{3}\cos^{3/2}x\;|_0^{\pi/2}\]
But yes, you're right, our cosines are going to be producing a bunch of 0's and 1's.

- anonymous

don't you have to change the limits?

- zepdrix

If we were going to evaluate this while it was in u, then yes.
\[\Large \frac{2}{7}\cos^{7/2}x-\frac{2}{3}\cos^{3/2}x\;|_{\color{green}{x=0}}^{\color{green}{\pi/2}}\]
Or,
\[\Large \frac{2}{7}u^{7/2}-\frac{2}{3}u^{3/2}\;|_{u=1}^{0}\]

- zepdrix

When you do a substitution, you have 2 options:
1. You can change the limits along with doing your substitution.
2. You can undo your substitution at the end, before you plug in the limits, in which case you don't need to change the limits.

- anonymous

oh, nice. That's part of what has been confusing me. So after I change it back from the u I can just plug in the original limits?

- zepdrix

ya c:

- zepdrix

You have the option to NOT CHANGE IT BACK TO U,
that's the case where you would have to figure out new limits.
Both methods work just fine :o

- anonymous

awesome, this problem cleared up a lot of confusion. I got 8/21 for the final answer

- zepdrix

yay good job \c:/

- anonymous

awesome! Thank you so much!

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