How to integrate sin(3x^2)?

shizz

The full question is 2xsin(3x^2). I've been trying to do it with by parts but I'm not sure how to integrate Sin(3x^2). When I put it into an online integrator i get an answer involving square roots and pi's, but when I put the full question straight into the integrator i get -(1/3)cos(3x^2) which makes me think you some how differentiate the (3x^2) and divide the 2x by it. I can get to this by using by parts, ie. the uv equals this but i dunno how to get rid of the vdu part.

Can anyone help me shed some light on this?

sponsoredwalk

Watch this for a bit of intuition:

https://www.youtube.com/watch?v=LxtxQgJcqNw

But it doesn't show everything so this after:

https://www.youtube.com/watch?v=qclrs-1rpKI

shizz

sponsoredwalk wrote: »

Watch this for a bit of intuition:

https://www.youtube.com/watch?v=LxtxQgJcqNw

But it doesn't show everything so this after:

https://www.youtube.com/watch?v=qclrs-1rpKI

Thanks man. Ill watch them later can't now. Much appreciated.

shizz

ah think it just occurred o me what to do. I though in these sort of situations, Ie function of x by a function of x, is by parts all the time?

sponsoredwalk

This problem is a "u du substitution" (or "u substitution") problem.
Usually people would see if a u-substitution would work on a problem
before considering integration by parts but even then there's no general
rule.

If I integrate ∫2x³dx normally I'll get ½x⁴ + C. Well if I break ∫2x³dx up
into ∫2x(x²)dx I can use the u-substitution method. By setting u = x² I
can differentiate this to get du/dx = 2x, or du = 2xdx. Now we see that
∫2x(x²)dx = ∫udu, which is easy to integrate: ∫udu = ½u² + C = ½x⁴ + C.
Notice what 2x(x²) looks like, it looks something like the chain rule, i.e.
2x(x²) = 2x(x)², you know - something inside brackets raised to a power
& something outside it as well. Basically a u-substitution is like a reversal
of the chain rule. Try to spot the similarities between this example & your
own one & give the videos a shot when you can.

LeixlipRed

To summarise, if you have to integrate a function of the form g(f(x).f'(x) then you let u=f(x) and apply the method as showed by Sponsored Walk there. What does that mean in English? i.e. if there's some function and it's derivative (or a multiple of it) contained within the expression you can use the method. Note, the derivative must be multiplied by dx, it can't be in the denominator of a fraction for example.

By parts works for products of functions. Which you can hopefully transform into something easier to integrate. Try using by parts here and I think you'll run into difficulty with integrating the sin(x^2) term.

shizz

Thanks guys I know how to do it now. It's funny, if I was shown that question before learning by parts, I would have been able to do it. Just after doing by parts I have it in my head that a function by a function is always by parts. Obviously this isn't the case. Thanks for your help.