How can the integral of cosine squared be differentiated with respect to x?

[azatkgz]

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]

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:
\frac{d}{dx}\int_{a(x)}^{b(x)}f(t)dt=\frac{d}{dx}(F(b(x))-F(a(x)))=F&#039;(b(x))b&#039;(x)-F&#039;(a(x))a&#039;(x)=f(b(x))b&#039;(x)-f(a(x))a&#039;(x)

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

 

[D H]

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:
<br /> \frac{d}{dx}\int_{x^3}^{e^x}\cos t^2dt =<br /> \frac{d}{dx}\int_0^{e^x}\cos t^2dt \;\;-\;\;<br /> \frac{d}{dx}\int_0^{x^3}\cos t^2dt

 

[azatkgz]

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

 

[D H]

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

 

[azatkgz]

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

 

[arildno]

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

 

[azatkgz]

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

 

[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 e^{2x},I typed e^{x}
\frac{d}{dx}(e^{2x})=2xe^{2x}
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
e^xcos(e^{2x})-3x^2cos(x^6)?

 

[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.

 

[arildno]

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

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

 

[azatkgz]

Ok.Thanks to everyone.