# 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.

Gib Z
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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
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Differentiate it, don't try to integrate it!

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

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

Last edited:
$$2e^xsine^{2x}-6x^5sinx^6$$
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
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Eeh??
Where do you get that 2-factor from??

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

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

$$e^xcos(e^{2x})-3x^2cos(x^6)$$?

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
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arildno
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$$e^xcos(e^{2x})-3x^2cos(x^6)$$?