intergral of sin function

[-GC-Phoenix]

OK well Im trying to figure out the intergral of the sin function below. So far I have no ideas of what the answer is, and neither does my calc teacher. Any help would be great.

 

[galactus]

That's because this isn't doable by elementary means.

It's what is known as a Sine Integral. Denoted by 'Si'.

\(\displaystyle \L\\Si=\int_{0}^{x}\frac{sin(t)}{t}dt\)

When I ran this through Maple, I got:

\(\displaystyle \L\\Si(\frac{\pi}{2}e^{x})\)

Here's a graph of the sine integral from 0 to 1. I don't know how you would express an indefinite integral for this, but if you have limits of integration, I would use a Riemann sum.

Google Sine Integral and you will find something on it.

 

[-GC-Phoenix]

let me rewrite this a bit, heres the actual problem as it was writen.

If f'(x) =

and f(0) = 1, then f(2) = ...


I was wanting the equation of the integral so I could solve that problem.

 

[Count Iblis]

let me rewrite this a bit, heres the actual problem as it was writen.

If f'(x) =

and f(0) = 1, then f(2) = ...


I was wanting the equation of the integral so I could solve that problem.

You can only calculate the integral from minus infinity to infinity. You first substitute x = Log(t) to obtain the integral Galactus gave above:

\(\displaystyle \L\int_{-\infty}^{0}\frac{\sin\left(\frac{\pi}{2}t)}{t}dt\)

You can remove the factor \(\displaystyle \frac{\pi}{2}\) by rescaling t. To calculate the integral, consider this integral first:

\(\displaystyle \L\int_{0}^{\infty}\sin(t)\exp(-s t)dt=\frac{1}{s^{2}+1}\)

This is easy to derive. Now you integrate both sides w.r.t. s from zero to infinity. On the left hand side you interchange the s and t integrations and evaluate the integral over s of the integrand. The integral over t then becomes the desired integral.

 

[skeeter]

I've seen problems like this before, and I believe they are meant to be solved using technology (a calculator) ...

the main idea behind the problem is the use of the fundamental theorem rather than trying to find an antiderivative of f' ... in fact, the functions for f' in these problems are often chosen such that they do not yield an elementary closed-form antiderivative.

\(\displaystyle \L f(2) - f(0) = \int_0^2 \sin{\left(\frac{\pi e^x}{2}\right)} dx\)

\(\displaystyle \L f(2) = f(0) + \int_0^2 \sin{\left(\frac{\pi e^x}{2}\right)} dx\)

\(\displaystyle \L f(2) = 1 + \int_0^2 \sin{\left(\frac{\pi e^x}{2}\right)} dx\)

calculate the right side of the equation to get f(2) = 1.157

 

[-GC-Phoenix]

thanks for your help guys.
Im thinking that what skeeter said is correct, because its the only one that provided a result that was given as a choice.

And because this is for a calc 1 course, and the other stuff isnt part of what we learn, but thanks anyways.