Question

Help please!
Find dy/dx.
Problems below.

Answer 1

\[1.) y= \int\limits_{0}^{x^2} (\cos \sqrt{t}) dt\]
\[2.) y=\int\limits_{\tan x}^{0} \frac{ dt }{ 1+t^2 }\]

Answer 2

have you learned substitution of variable trick?

Answer 3

I've learned substitution but have a hard time applying it, and doing it. @katherine.ok

Answer 4

So, for a, you will let u= y^(1/2) then you will get du in terms of dt.

Answer 5

u=t^(1/2)**

Answer 6

okat

Answer 7

No actually, this doesnt get you anywhere. LOL

Answer 8

I tried, and it doesnt simplify at all.

Answer 9

\[du= \frac{ 1 }{ 2\sqrt{x} }\]

Answer 10

@katherine.ok

Answer 11

@Loser66

Answer 12

I got 2 integral (ucosudu)

Answer 13

wait how?

Answer 14

if you let
\[u=t^{1/2}\]
then\[du=\frac{ 1 }{ 2\sqrt{x} }\]

Answer 15

is (u) not t^1/2?
@katherine.ok

Answer 16

\[\int\limits_{?}^{?}fdg=fg-\int\limits_{?}^{?}gdf\]

Answer 17

that confused me haha

Answer 18

you know that f= u, dg must be cosu du, df=du, g= sinu

Answer 19

The above is just a theorem of integration. just remember it. :((

Answer 20

you got du=(1/2)t^(-1/2)dt,so solve for dt, and thats how you get integral 2ucosudu

Answer 21

that makes no sense :/ @katherine.ok

Answer 22

could you show me work as to how you got it.

Answer 23

u=t^(1/2), du= (1/2)t^(-1/2)dt, dt= 2t^(1/2)du, substituting it into cos t^(1/2)du= you get

Answer 24

dt**

Answer 25

integral 2ucosudu

Answer 26

take the two outside of integral sign as it is constant--> 2integral ucosudu

Answer 27

I think katherine.ok (hehe, this nick is ok, then) is right. from here take integral by part.

Answer 28

I wasn't sure if he would know integration by parts. I just wanted to sit back just in case there was some trick to this that didn't require by parts xD

Answer 29

Well, without knowing myself, I have a50/50 shot, right? xD @Loser66

Answer 30

Or you could just say you win at being a loser? O.o

Answer 31

So any idea about integration by parts? :P

Answer 32

this question was done by integration by parts...

Answer 33

Well, I'm used to seeing this kind of question in a calc 1 class, so didn't know if there was maybe another way then. Sorry.....

Answer 34

I got 2(sqrt(t)sint^(1/2)+cost^(1/2))+ Constant

Answer 35

\[u= t^{1/2}\]
\[du= \frac{ 1 }{ 2 } t^{-1/2}\]
\[dt=2t^{1/2}\]
\[=> cost^{1/2} dt\]
\[=\int\limits_{}^{} 2u*\cos(u) du = 2\int\limits_{}^{} u*\cos(u) du\]

Answer 36

YAY that is correct.

Answer 37

Definite integral, no + C :P

Answer 38

Now pdd you need to do
∫??fdg=fg−∫??gdf, where u=t^(1/2), du= (1/2)t^(-1/2)dt, dt= 2t^(1/2)du,

Answer 39

ok. and yes +C @Psymon
where do I go from there to get dy/dx @katherine.ok

Answer 40

Your limits are from 0 to x^2, though, why would there be a + C?

Answer 41

∫fdg=fg−∫gdf

Answer 42

It doesnt matter if theres a c or not. I always do indefinite first, then sub in values.

Answer 43

nvm, it has an interval so no constant. @Psymon you confused me.

Answer 44

how do i evaluate the integral with ∫fdg=fg−∫gdf

Answer 45

And when you sub in, constant term can be ignored. right?

Answer 46

I was just remarking because she initially put + C. I was correcting just in case......

Answer 47

ahhhhhh! i'm getting confused and frustrated -.-

Answer 48

f= u, dg must be cosu du, df=du, g= sinu

Answer 49

Just do by parts, don't worry about c or no c yet :P

Answer 50

got the first part. so now how do i evaluate it. and I didnt learn what you were trying to describe @katherine.ok

Answer 51

∫fdg=fg−∫gdf, where f= u, dg must be cosu du, df=du, g= sinu... soooo fg= usinu , ∫gdf is ∫sinudu, and now you can INTEGRATE THIS

Answer 52

Have you ever seen integration by parts @pdd21 ?

Answer 53

no i have not. @Psymon that's my problem... if I show my work @katherine.ok 's way my professor will be confused as to how I know this..

Answer 54

After that integration of ∫gdf, and finding what fg−∫gdf is equal to, substitute min and max, and there is your answer.

Answer 55

Yeah, that's my concern, too. That being said, are you sure this isn't finding the derivative of the integral? Sinceyou havent seen by parts before.

Answer 56

This is integration by parts combined with ∫fdg=fg−∫gdf formula. Nothing special...

Answer 57

it says to find dy/dx......

Answer 58

but i didn't learn it that way. @katherine.ok

Answer 59

Can't be trying to show him something in calc 2 when she's in calc 1 xD

Answer 60

exactly what @Psymon said @katherine.ok

Answer 61

After integration, you will have y= numbers and x's. then you get dy/dx... I dont know simpler way than the way i presented to you sorry:(

Answer 62

Yeah, I only know the integration by parts method, too. If this were the derivative of theintegral, I would know what to do, but without that I am unsure. I doubt you have any answer choices, huh? xD

Answer 63

No i dont have answer choices. ;/ @Psymon

Answer 64

could you show me the integration by parts...

Answer 65

Alright. Well, I use different letters than katherine uses, but this is theidea. You usually have two things you can pick in your integral. One of the things you choose must be easy or convenient to differentiate and the 2nd thing must be easy or convenient to integrate. Now it may not look like it, but you do have 2 good candidates. What I would choose to differentiate would be cos(t^(1/2)) and what I could choose to integrate would be simply dt. Now what we choose to differentiate I would call u and what I choose tointegrate I would call v. Make sense so far?

Answer 66

yes.

Answer 67

Alright, cool. Now the formula for integration by parts is this then:
\[u \int\limits_{}^{}v - [u'\int\limits_{}^{}v]\]
So essentially we take regular u and multiply it by integral v then subtract what we get after doing u prime times integral v.

Answer 68

v is the funtion right?

Answer 69

Well, u and v are pieces of the entire function. In this case, I chose u to be cos(t^(1/2)) and v to be dt. I chose u to be the cos term because in the integration by parts formula, you need one term to differentiate and one term to integrate. cos is the easy term to differentiate and dt is the easy term to integrate.

Answer 70

oh okay..

Answer 71

Mhm. So what would be integral dt?

Answer 72

\[\cos(t^{1/2})\int\limits_{}^{} dt - [ v' \int\limits_{}^{} dt\]

Answer 73

Right. we just need to know what u' and integral v are, though :P integral dt is just t. Think you got the derivative of cost^1/2?

Answer 74

\[u'=\frac{ -\sin \sqrt{t} }{ 2\sqrt{t} }\]

Answer 75

@Psymon

Answer 76

Sorry, had to be away for a sec. But yeah, that would be your derivative. I'm just jumping ahead and checking the solution myself is all.

Answer 77

so..
\[\cos(t^{1/2})\int\limits_{}^{} dt = \frac{ -\sin \sqrt{t} }{ 2 \sqrt{t} } \int\limits_{}^{} dt\]

Answer 78

where do I go from there? @Psymon

Answer 79

Hmm...it looks like we do need to do that u-substitution first -_-

Answer 80

But the method is still the same. Just was looking at it and it looks like it goes in circles if you don't sub first like katherine did.

Answer 81

Bleh, guess I suck at these more than I think. Alright, so we'll do integration by parts with what katherine had
\[2\int\limits_{}^{}u cosu du \] so I'll just choose my u (what Ill take the derivative of) to be u and my v (what Ill integrate) to be cosu.
So back to the formula, which apparently I typed wrong *headdesk*
\[u \int\limits_{}^{}v - [\int\limits_{}^{}(u'\int\limits_{}^{}v)]\]
It's a bit funky, but I'll plug in the values and you can see. So since my....u was u, u' is just 1. And since my v was cosu, my integral v is sinu. So now I'll plug all this into the formula to get:
\[usinu -[\int\limits_{}^{}(1sinu)]\] So basically in that second part, you do two integrals. The integral of v and then the integral of u' times integral v. So if I simplify this I then have:
\[usinu + cosu\] Since u = sqrt(t), I'll sub that back in.

\[\sqrt{t}*\sin(\sqrt{t}) + \cos (\sqrt{t})\] from 0 to x^2.

Answer 82

Oh, and forgot to multiply the 2 back in x_x

Answer 83

\[2\sqrt{t}*\sin(\sqrt{t}) + \cos(\sqrt{t})\]

Answer 84

ohhhhh... I see that makes more sense now.

Answer 85

Yeah, lol. It's just I forgot to add the extra integral in the 2nd part. But yes, you pick something easy to differentiate and something easy to integrate. Then the combination of the two using that formula gets you your integral. Now just plug in your limits :3

Answer 86

2nd one is WAYYYYY easier o.o

Answer 87

@SithsAndGiggles Yo, since you're here. You know a way to do her first integral problem without doing by parts?

Answer 88

As far as I can tell, there's no integration that needs to be done. The problem says "Find dy/dx." Think fundamental theorem of calc:
\[\frac{d}{dx}\int_c^{g(x)}f(t)~dt=f(g(x))\times g'(x)\]

Answer 89

Right, I would've thought that as well if it was what you posted d/dx of the integral. But since I didn't see that, I wasn't sure.

Answer 90

But just to entertain the idea of integrating, a substitution should do the trick:
\[\int_0^{x^2}\cos\sqrt t~dt\]
Let \(u=\sqrt t~\Rightarrow~du=\dfrac{1}{2\sqrt t}~dt~\Rightarrow~2u~du=dt\).
\[2\int_0^{x}u\cos u~du\]
Followed by integration by parts.

Answer 91

...which @katherine.ok suggested earlier.

Answer 92

Right. And which we started going off of. I'm just too stupid to try and sub first before doing integration by parts. Once I started running in circles I just went back to what katherine did. But I was mainly curious to see if there was something else. I would definitely think this is fundamental theorem, but wasnt positive.

Answer 93

Yeah, like I said before, this is definitely a problem of applying FTC since you're differentiating an integral.

So for the first integral, if
\[y=\int_0^{x^2}\cos\sqrt t~dt\]
then its derivative with respect to x is
\[\begin{align*}\frac{dy}{dx}&=\frac{d}{dx}\int_0^{x^2}\cos\sqrt t~dt\\
&=\cos\left(x^2\right)\times 2x\\
&=2x\cos\left(x^2\right)
\end{align*}\]
And that would be the answer to the first @pdd21.

Answer 94

okay. that made it easier to understand @SithsAndGiggles

Answer 95

Dunno why my brain didn't think differentiating it x_x I start doing these problems and realize I don't know them as well as I think.