Differentiation of Integral

azatkgz

1. The problem statement, all variables and given/known data


[tex]\frac{d}{dx}\int_{x^3}^{e^x}cost^2dt[/tex]



3. The attempt at a solution

[tex]\int cost^2dt=\frac{sint^2}{2t}+\int\frac{sint^2}{2t^2}dt[/tex]
[tex]\int\frac{sint^2}{2t^2}dt=-\frac{sint^2}{2t}+\int cost^2dt[/tex]
I came back to initial integral.

Gib Z

That can happen with some of your choice for u and dv whilst doing integration by parts, the second time you apply it use different choices.

arildno

Differentiate it, don't try to integrate it!

Gib Z

I think he needs to evaluate the integral to be able to do that doesn't he >.<

arildno

Nope.
Here's how to do it properly:
Let F(t) be an antiderivative of f, F'(t)=f(t).
Thus, we have:
[tex]\frac{d}{dx}\int_{a(x)}^{b(x)}f(t)dt=\frac{d}{dx}(F(b(x))-F(a(x)))=F'(b(x))b'(x)-F'(a(x))a'(x)=f(b(x))b'(x)-f(a(x))a'(x)[/tex]

As you can see, you do not need the explicit form of F, only the guarantee that some such F exists..

D H

No. The integrand does not involve x. Simply apply the fundamental theorem of calculus.

Hint:
[tex]
\frac{d}{dx}\int_{x^3}^{e^x}\cos t^2dt =
\frac{d}{dx}\int_0^{e^x}\cos t^2dt \;\;-\;\;
\frac{d}{dx}\int_0^{x^3}\cos t^2dt[/tex]

azatkgz

So tha answer is
[tex]2e^xsine^{2x}-6x^5sinx^6[/tex]
yes?

D H

No. Read arildno's post again. Sine is not involved.

azatkgz

ok ok,my mistake
[tex]2e^{2x}cos(e^{2x})-6x^5cos(x^6)[/tex]

arildno

Eeh??
Where do you get that 2-factor from??

azatkgz

if we put [tex]e^x [/tex] to t shouldn't it be [tex]e^{2x}[/tex]

arildno

I'm talking about the 2-factors in front of the cosine's, not the ones within the arguments.

azatkgz

I typed wrongly instead of [tex]e^{2x}[/tex],I typed [tex]e^{x}[/tex]
[tex]\frac{d}{dx}(e^{2x})=2xe^{2x}[/tex]
This 2 are you asking ?

arildno

What is a(x), and what is b(x); what are their derivatives?

azatkgz

You say that answer is
[tex]e^xcos(e^{2x})-3x^2cos(x^6)[/tex]?

Gib Z

Today was not my best day obviously =] Yes I should have seen the proper method arildno and DH, maybe Ill have better luck tomorrow.

HallsofIvy

No, no one has said that! Several people have asked you questions about this problem that you haven't answered.