Question

How would I put cos(u^2) in to terms of x using u sub. Could I....

Answer 1

say u=x^2 and du= 2x dx and then say that the integral cos(u^2)du is equal to cos(x)*2x dx

Answer 2

switch that u^2=x

Answer 3

I'm not sure I follow what you're trying to do? Typically you'll have something like
\[\int u\ cos(u^2)\ du\]

Answer 4

the question is integrate cos(u^2)du from cos(x) to 5x

Answer 5

My mistake, Find the derivative of the following y= the integral from cos(x) to 5x of cos(U^2)du

Answer 6

Oh, whew.. Much better ;p

Answer 7

So I assume you put u into terms of x, then integrate, then evaluate, then take the derivative?

Answer 8

Are you taking the derivative with respect to x or with respect to u?

Answer 9

x I think.

Answer 10

It doesn't say?

Answer 11

It should say.

Answer 12

I will put in the equation as stated

Answer 13

Find the derivative of the following function: \[y=\int\limits_{cosx}^{5x} \cos(u^2)du\]

Answer 14

so I decided to let x=u^2 and du=2x so the equation becomes integral of cos(x)*2xdx

Answer 15

I don't think you need to do that. I feel as though they want you to rely on the fundamental theorem of calculus here..

Answer 16

Wait so I can just take the integral of cos(u^2) and then eval for 5x and cos x?

Answer 17

and then take the derivative of the evaluated function?

Answer 18

Oh, wait, sure you can.. just requires a u-sub, then parts. Sorry.

Answer 19

\[\int\limits 2xcos(x) dx \rightarrow 2xsin(x)+2\cos(x)\]

Answer 20

I dunno where to go with it. The answer is sin(x)*cos(cos^2x)+5cos(25x^2)

Answer 21

Ah forgot to change the limits.

Answer 22

You have to do parts again on 2xsin(x) so it repeats and gets you back to 2xcos(x)

Answer 23

It doesn't repeeat does it U=2x du =2 v=sin(x) dv = cos(x)

Answer 24

Oh, you're right.

Answer 25

Wait does the question actually ask? \[d/du \int\limits_{\cos(x)}^{5x} \cos(u^2)du\]

Answer 26

Do the integral and the derivative cancel?

Answer 27

Wait, what did you use for the original u-sub?

Answer 28

u^2=x

Answer 29

\[u^2 = x \implies 2u\ du = dx \implies du = \frac{1}{2\sqrt{x}}dx\]

Answer 30

So I don't think you're using the right integral.

\[\int cos(u^2)du = \int {cos(x) \over 2\sqrt{x}}dx\]

Answer 31

Probably I was using du=2xdx

Answer 32

I can see why it would be 1/2x I don't see how it is root x

Answer 33

Because \(u^2 = x \implies u = \sqrt{x}\)

So
\[2u\ du = dx \implies du = \frac{1}{2u}dx = \frac{1}{2\sqrt{x}}dx\]

Answer 34

I think we need to use the fundamental theorem. And I think it'll make this much easier..

Answer 35

Oh I see.

Answer 36

Lets see here. Lets assume that \[F(u) = \int cos(u^2) du\]
\[\implies \int\limits_{\cos(x)}^{5x} \cos(u^2)du = F(5x) - F(cos(x))\]
Right?

Answer 37

Right

Answer 38

Ok, well if \[F(u) = \int cos(u^2) du\]
Then\[\frac{d}{du}F(u) = cos(u^2) \]

Answer 39

Right?

Answer 40

I see so the integral and the derivative cancel?

Answer 41

differential*

Answer 42

Yes. That's the fundamental theorem of calculus. Integration and derivatives are inverses of eachother (plus or minus a constant).

Answer 43

So the question is just asking cos(5x^2)-cos(cos(x^2)

Answer 44

So that means that
\[\frac{d}{dx} \int\limits_{\cos(x)}^{5x} \cos(u^2)du =\frac{d}{dx}[ F(5x) - F(cos(x))]\]\[= cos((5x)^2)( \frac{d}{dx}[5x]) - cos(cos^2(x))(\frac{d}{dx}[cos(x)])\]\[=?\]

Answer 45

Oh snap. 5cos(25x^2)+sin*cos(cos^2(x)) which is the answer.

Answer 46

So just have to remember that the derivative undoes the integral, then be sure to use the chain rule since your limits are functions of another variable.

Answer 47

That's it great thanks.