J jbstahley New member Joined Jan 28, 2007 Messages 3 Jan 28, 2007 #1 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
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
G galactus Super Moderator Staff member Joined Sep 28, 2005 Messages 7,203 Jan 28, 2007 #2 Let \(\displaystyle \L\\u=x^{2}, \;\ du=2xdx, \;\ \frac{1}{2}du=dx\) \(\displaystyle \L\\\frac{1}{2}\int_{0}^{\pi^{2}}[cos(u-{\pi}^{2})]du=\frac{1}{2}|_{0}^{\pi^{2}}sin(u-{\pi^{2}})=\frac{sin({\pi^{2}})}{2}\)
Let \(\displaystyle \L\\u=x^{2}, \;\ du=2xdx, \;\ \frac{1}{2}du=dx\) \(\displaystyle \L\\\frac{1}{2}\int_{0}^{\pi^{2}}[cos(u-{\pi}^{2})]du=\frac{1}{2}|_{0}^{\pi^{2}}sin(u-{\pi^{2}})=\frac{sin({\pi^{2}})}{2}\)
S soroban Elite Member Joined Jan 28, 2005 Messages 5,584 Jan 29, 2007 #3 Re: Definite Integral Hello, jbstahley! It's straight substitution . . . \(\displaystyle \L\int^{\;\;\;\pi}_0 x\cdot\cos(x^2\,-\,\pi^2)\,dx\) Click to expand... 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: \(\displaystyle \L\:\frac{1}{2}\sin(\pi^2\,-\,\pi^2)\,-\,\frac{1}{2}\sin(0^2\,-\,\pi^2)\) . . \(\displaystyle \L= \;\frac{1}{2}(0)\,-\,\frac{1}{2}\sin(-\pi^2)\) . . \(\displaystyle \L=\:-\frac{1}{2}\sin(-\pi^2)\) . . \(\displaystyle \L=\:-0.215150609\)
Re: Definite Integral Hello, jbstahley! It's straight substitution . . . \(\displaystyle \L\int^{\;\;\;\pi}_0 x\cdot\cos(x^2\,-\,\pi^2)\,dx\) Click to expand... 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: \(\displaystyle \L\:\frac{1}{2}\sin(\pi^2\,-\,\pi^2)\,-\,\frac{1}{2}\sin(0^2\,-\,\pi^2)\) . . \(\displaystyle \L= \;\frac{1}{2}(0)\,-\,\frac{1}{2}\sin(-\pi^2)\) . . \(\displaystyle \L=\:-\frac{1}{2}\sin(-\pi^2)\) . . \(\displaystyle \L=\:-0.215150609\)
