Use a valid integration method to calculate $\int_{\pi/2}^{3\pi/2}\frac{\sin x}{x}\,dx$

Question

We know that $f'(x)$ = $\dfrac{\sin x}{x}$ and $f(\pi/2)=0$, $f(3\pi/2)=1 $ Use a valid integration method to evaluate : $$\int_{\pi/2}^{3\pi/2}f(x)dx$$

i think $\int_{\pi/2}^{3\pi/2}\frac{\sin x}{x}\,dx $ is the $f(x)$ function we're looking for, i used barrow theorem but i don´t know how to integrate it, any helps or hint will appreciate.

Answer 1

Thinking and trying, i got the solution :

$\int_{\pi/2}^{3\pi/2}f(x)dx $

Now we can use integration by parts, by it's definition , $u=f(x)$, $du=f'(x)dx$ and $dv=dx$, $v=x $,by ILATE rule :

\begin{align} = f(x) x|_{\pi/2}^{3\pi/2}-\int xf'(x)dx \\ & =\require{cancel} [\cancelto{1}{f(3\pi/2)}(3\pi/2)]-[\cancelto{0}{f(\pi/2)}(\pi/2)]-\int_{\pi/2}^{3\pi/2}\cfrac {\cancel{x}\ cosx}{\cancel{x}} \\ & = \Bigl(\cfrac{3\pi}{2}-0\Bigr) -\ sen x|_{\pi/2}^{3\pi/2} \\ & = \cfrac {3\pi}{2}-[(-1)-1] \\ & = \cfrac {3\pi}{2}+2 \\ \end{align}

Answer 2

You can't.

$\dfrac{\sin(x)}{x} $, also called $sinc(x) $ (sometimes with a $\pi$), is not integrable in finite terms.

It has been given a name: $Si(x) =\int_0^x \dfrac{sin(t)}{t}dt $.

Special forms, such as $\int_0^{\infty}$ can be evaluated.

Your case can only be done numerically.

Wolfy says $Si((3 π)/2) - Si(π/2)≈0.23761058579952225576304433203202186890796018928024067084207168096967568361245277161035797838127400 $

Answer 3

If your function is $f(x)=\frac{\sin(x)}{x}$, and your task is to calculate $$\int_{\pi/2}^{3\pi/2}f'(x)\mathrm{d}x$$ Then you can use the Fundamental Theorem of Calculus to calculate the integral: $$\int_{\pi/2}^{3\pi/2}f'(x)\mathrm{d}x=f\left(\frac{3\pi}{2}\right)-f\left(\frac{\pi}{2}\right)$$