Antiderivative of x^2sin(x^2) with Given Initial Condition

[Xcron]

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.

Any advice/help?

 

[d_leet]

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.
Any advice/help?

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 $$x\sin(x^2)$$

 

[Xcron]

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:
$$x^2\sin(x^2)=x^2\sin(x^2)(\sin^2(x^2)+\cos^2(x^2))=<br /> x^2\sin^3(x^2)+x^2\sin(x^2)\cos^2(x^2)$$

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]

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.

 

[dextercioby]

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

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

Daniel.