# Differentiation of Integral

## Homework Statement

$$\frac{d}{dx}\int_{x^3}^{e^x}cost^2dt$$

## The Attempt at a Solution

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

## Answers and Replies

Gib Z
Homework Helper
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
Homework Helper
Gold Member
Dearly Missed
Differentiate it, don't try to integrate it!

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

arildno
Homework Helper
Gold Member
Dearly Missed
Nope.
Here's how to do it properly:
Let F(t) be an antiderivative of f, F'(t)=f(t).
Thus, we have:
$$\frac{d}{dx}\int_{a(x)}^{b(x)}f(t)dt=\frac{d}{dx}(F(b(x))-F(a(x)))=F'(b(x))b'(x)-F'(a(x))a'(x)=f(b(x))b'(x)-f(a(x))a'(x)$$

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

D H
Staff Emeritus
I think he needs to evaluate the integral to be able to do that doesn't he >.<
No. The integrand does not involve x. Simply apply the fundamental theorem of calculus.

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

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So tha answer is
$$2e^xsine^{2x}-6x^5sinx^6$$
yes?

D H
Staff Emeritus
No. Read arildno's post again. Sine is not involved.

ok ok,my mistake
$$2e^{2x}cos(e^{2x})-6x^5cos(x^6)$$

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arildno
Homework Helper
Gold Member
Dearly Missed
Eeh??
Where do you get that 2-factor from??

if we put $$e^x$$ to t shouldn't it be $$e^{2x}$$

arildno
Homework Helper
Gold Member
Dearly Missed
I'm talking about the 2-factors in front of the cosine's, not the ones within the arguments.

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

arildno
Homework Helper
Gold Member
Dearly Missed
What is a(x), and what is b(x); what are their derivatives?

You say that answer is
$$e^xcos(e^{2x})-3x^2cos(x^6)$$?

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

arildno
Homework Helper
Gold Member
Dearly Missed
You say that answer is
$$e^xcos(e^{2x})-3x^2cos(x^6)$$?

I didn't say that; but you said it correctly now!

Ok.Thanks to everyone.