Definite Integral: int[ x*Cos[x^2 - Pi^2], x] from 0 to PI

[jbstahley]

I really could use help on this definite integral. Help is very appreciated.

Integrate[ x*Cos[x^2 - Pi^2], x] from 0 to PI

 

[galactus]

Let \(\displaystyle \L\\u=x^{2}, \;\ du=2xdx, \;\ \frac{1}{2}du=dx\)

\(\displaystyle \L\\\frac{1}{2}\int_{0}^{\pi^{2}}[cos(u-{\pi}^{2})]du=\frac{1}{2}|_{0}^{\pi^{2}}sin(u-{\pi^{2}})=\frac{sin({\pi^{2}})}{2}\)

 

[soroban]

Re: Definite Integral

Hello, jbstahley!

It's straight substitution . . .

\(\displaystyle \L\int^{\;\;\;\pi}_0 x\cdot\cos(x^2\,-\,\pi^2)\,dx\)

Let \(\displaystyle u \:=\:x^2\,-\,\pi^2\;\;\Rightarrow\;\;du\:=\:2x\,dx\;\;\Rightarrow\;\;dx\:=\:\frac{du}{2x}\)

Substitute: \(\displaystyle \L\:\int x\cdot\cos u \left(\frac{du}{2x}\right)\:=\:\frac{1}{2}\int \cos u\,du \:=\:\frac{1}{2}\sin u\)


Back-substitute: \(\displaystyle \:\frac{1}{2}\sin\left(x^2\,-\,\pi^2\right)\,\L |^{\pi}_0\)

Evaluate: \(\displaystyle \L\:\frac{1}{2}\sin(\pi^2\,-\,\pi^2)\,-\,\frac{1}{2}\sin(0^2\,-\,\pi^2)\)

. . \(\displaystyle \L= \;\frac{1}{2}(0)\,-\,\frac{1}{2}\sin(-\pi^2)\)

. . \(\displaystyle \L=\:-\frac{1}{2}\sin(-\pi^2)\)

. . \(\displaystyle \L=\:-0.215150609\)