anonymous
  • anonymous
integrate cos^(1/2)sin^3x dx from 0 to pi/2
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
zepdrix
  • zepdrix
\[\Large \int\limits_0^{\pi/2}\sqrt{\cos x}\;\sin^3x\;dx\]So this is what we've got to solve? Hmm
anonymous
  • anonymous
Yes, so far I've pulled out one of the sinx then changed the \[\sin ^{2}x \to (1-\cos ^{2}x)\]
anonymous
  • anonymous
then I guess you do a u sub and u=cosx and du would be -sinx

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
but I'm not sure where to go from there
zepdrix
  • zepdrix
-du=sinx? looks like you're on the right track
zepdrix
  • 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
  • zepdrix
If I wrote it like this, does it become a tad clearer?\[\Large -du=(\sin x\;dx)\]
zepdrix
  • zepdrix
Maybe you can match it up with the problem that way hehe
anonymous
  • anonymous
ahh, I see what you're saying
anonymous
  • anonymous
then the negative on that would just come out from like a constant
zepdrix
  • 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
  • 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
  • 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
  • zepdrix
writing x= is a good way to remind yourself of that.
anonymous
  • anonymous
oh, okay. Yeah, I usually forget that :/
zepdrix
  • zepdrix
So what's the next step? \c:/
anonymous
  • anonymous
I would probably distribute the \[u ^{1/2}\] to the parenthesis then solve the equation like seperate integrals
zepdrix
  • zepdrix
sounds good :O
anonymous
  • anonymous
so now I have \[-(\frac{ 2 }{ 3 })u ^{3/2}-(\frac{ 1 }{ 3 })u ^{3}\]
anonymous
  • anonymous
or did I forget something?
zepdrix
  • 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
  • zepdrix
\[\Large u^{1/2}u^{2} \quad=\quad u^{5/2}\]
anonymous
  • anonymous
apparently I can do calculus but can't add :/ lol
zepdrix
  • zepdrix
hehe
anonymous
  • anonymous
but did I have the rest correct? then I'd sub back for the u?
zepdrix
  • zepdrix
AHHHH stupid openstudy freezing >:(
zepdrix
  • 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
  • anonymous
yeah :/
anonymous
  • anonymous
I can't complain too much though. Don't know what I'd do without it
zepdrix
  • zepdrix
From there, ya let's undo our substitution, then we can evaluate the result at our limits.
anonymous
  • anonymous
ah, okay
anonymous
  • anonymous
\[\left[ -\frac{ 2 }{ 3 } \right]cosx ^{3/4}+\frac{ 2 }{ 7 }\cos ^{7/2}\] evaluated at 1 to 0?
anonymous
  • anonymous
\[\left[ -\frac{ 2 }{ 3 } \right]cosx ^{3/4}+\frac{ 2 }{ 7 }\cos ^{7/2}\] evaluated at 1 to 0?
zepdrix
  • zepdrix
The first one is to the 3/2 power right? :o
anonymous
  • anonymous
yeah, typo
zepdrix
  • 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
  • anonymous
don't you have to change the limits?
zepdrix
  • 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
  • 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
  • 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
  • zepdrix
ya c:
zepdrix
  • 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
  • anonymous
awesome, this problem cleared up a lot of confusion. I got 8/21 for the final answer
zepdrix
  • zepdrix
yay good job \c:/
anonymous
  • anonymous
awesome! Thank you so much!

Looking for something else?

Not the answer you are looking for? Search for more explanations.