Question

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

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

but I'm not sure where to go from there

Answer 5

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

Answer 6

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

\[\Large u=\cos x \qquad\to\qquad -du=\sin x\;dx\]

Answer 7

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

Answer 8

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

Answer 9

ahh, I see what you're saying

Answer 10

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

Answer 11

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.

Answer 12

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

Answer 13

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

Answer 14

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

Answer 15

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

Answer 16

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

Answer 17

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

Answer 18

sounds good :O

Answer 19

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

Answer 20

or did I forget something?

Answer 21

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

\[\Large -\int\limits\limits\limits_{\color{red}{x=0}}^{\color{red}{\pi/2}} u ^{1/2}(1-u ^{2}) du \qquad\to\qquad \Large \int\limits\limits_{\color{red}{x=0}}^{\color{red}{\pi/2}} u ^{5/2}-u^{1/2}\; du\]

Answer 22

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

Answer 23

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

Answer 24

hehe

Answer 25

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

Answer 26

AHHHH stupid openstudy freezing >:(

Answer 27

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

Answer 28

yeah :/

Answer 29

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

Answer 30

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

Answer 31

ah, okay

Answer 32

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

Answer 33

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

Answer 34

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

Answer 35

yeah, typo

Answer 36

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.

Answer 37

don't you have to change the limits?

Answer 38

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,
\[\Large \frac{2}{7}u^{7/2}-\frac{2}{3}u^{3/2}\;|_{u=1}^{0}\]

Answer 39

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.

Answer 40

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?

Answer 41

ya c:

Answer 42

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

Answer 43

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

Answer 44

yay good job \c:/

Answer 45

awesome! Thank you so much!