# Antiderivative problem

Problem states:

Find and antiderivative $$F$$ of $$f(x)=x^2\sin(x^2)$$ such that $$F(1)=0$$.

I thought at first that I would use $$u=x^2$$ to simplify the matter but I don't seem to get anywhere. I can't seem to figure out a way to start it off either...I was considering the Squeeze Theorem because they are functions, essentially.

Xcron said:
Problem states:
Find and antiderivative $$F$$ of $$f(x)=x^2\sin(x^2)$$ such that $$F(1)=0$$.
I thought at first that I would use $$u=x^2$$ to simplify the matter but I don't seem to get anywhere. I can't seem to figure out a way to start it off either...I was considering the Squeeze Theorem because they are functions, essentially.

How do you use the squeeze theorem to find an antiderivative? Try integration by parts. If you haven't learned integration by parts yet, then are you sure it isn't $$x\sin(x^2)$$

Nope...sorry, I haven't learned integration by parts yet...that will be covered in the second Calculus class that I will be taking in this Winter session at my college.

And yes, I am sure it isn't $$x\sin(x^2)$$...that would be a completely superficial problem that I should not have any problems with =P.

I was thinking about this problem some more (been stuck on it since last night and it's almost the night of the next day haha) and I thought that I could possibly get an antiderivative that had $$\frac{1}{2}$$ in it but then I realized that no matter what I came up with...the product rule would ruin my plan. If only that $$x^2$$ was a constant haha...

About an hour ago, I thought about a method that I could use by adding in a few functions that wouldn't change the value of $$f(x)$$ and then I came up with a plan all of a sudden and it seemed to work in my head. I worked it over a few times in my head and realized that it may be the answer...I would have to use the law that states $$\sin^2(x)+\cos^2(x)=1$$.

Sadly, that didn't produce the answer either because I ended up with:
$$x^2\sin(x^2)=x^2\sin(x^2)(\sin^2(x^2)+\cos^2(x^2))= x^2\sin^3(x^2)+x^2\sin(x^2)\cos^2(x^2)$$

This didn't get me anywhere because the product rule caught me none the less...the $$x^2$$ was the problem in that case.

While typing all of this up, I decided to look in the back of the book to see what the answer is so that I could get some insight as to how to go about this problem....and I was a bit dumbfounded upon seeing it...I guess I forgot that an antiderivative can just be the general notation of an integral and the integrand...you don't necessarily need the actual antiderivative function...*sigh*...mmmm the answer was like this:

$$F(x)=\int_{1}^{x}t^2\sin(t^2)dt$$

HallsofIvy
Homework Helper
That happened to me in a differential equations course once- I reduced the problem to an integral and spent hours trying to do the integral. The answer in the back of the book gave the solution in terms of the integral!

Tx
This is probably a properties of definite integrals question, cause there is no way you can integrate this that I know of without using IBP.

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dextercioby
Homework Helper
$$\int x^{2}\sin x^{2}dx= -\frac{1}{2}\left( \cos x^{2}\right) x+\frac{1}{4}\sqrt{2}\sqrt{\pi }\func{FresnelC}\left( \frac{\sqrt{2}}{\sqrt{\pi }}x\right) + C$$

Well, set x=1 in the RHS and equal it to 0.

Daniel.