# Thread: Easy trig integral =/

1. ## Easy trig integral =/

I haven't done any sort of trig integrals in the past and i have pretty much forgotten all my identities since i haven't looked at them for over 3 years so please bear with me.

The question is:

upper limit = 1, lower limit = 0.
Integrate cos (x^2) with those bounds.

First off I had to approximate the integral using the midpoint/trapezoidal rules. Those were fairly straightforward. Now i want to check my answer but I'm kinda lost.

If i took the integral of cos (x^2), i would get sin(x^2) no?
Subbing in the limits of integration i would get sin(1) - sin(0) = sin (1)
but using the approximation rules above I got the integral as being approximately equal to one, which does not equal to sin (1).

Help?

2. Originally Posted by Kuma
I haven't done any sort of trig integrals in the past and i have pretty much forgotten all my identities since i haven't looked at them for over 3 years so please bear with me.

The question is:

upper limit = 1, lower limit = 0.
Integrate cos (x^2) with those bounds.

First off I had to approximate the integral using the midpoint/trapezoidal rules. Those were fairly straightforward. Now i want to check my answer but I'm kinda lost.

If i took the integral of cos (x^2), i would get sin(x^2) no? Mr F says: No. Differentiate it and you will see why.
Subbing in the limits of integration i would get sin(1) - sin(0) = sin (1)
but using the approximation rules above I got the integral as being approximately equal to one, which does not equal to sin (1).

Help?
The integral of cos(x^2) cannot be found in closed form using elementary functions.

3. Thanks. But how would i differentiate that then? Does chain rule apply or something. I've never worked with trig functions before so. I do know that d/dx cos x = sin x, how would that apply to cos x^2?

4. Yeah, you use the chain rule. First, find the derivative of sin, then find the derivative of the value within sin.

5. Ah i see. So you can differentiate cos x^2 to get 2x sin x^2, but the integral of that would pretty much be impossible to find. Makes sense.

6. I find that $\displaystyle \int_{0}^{1}\cos{x^2}\;{dx} = \sum\limits_{k=0}^{\infty}\frac{(-1)^k}{(2k)!(4k+1)}$.
(Not that it's significant or matters, anyway).