\int\limits^ (1)/(2)_0 x cos( \pi x), dx

Question

`\int\limits^ (1)/(2)_0 x cos( \pi x), dx`

Answer 1

Set
`y=\pi x`, so that
`\frac{\mathrm dy}\pi=\mathrm dx` and the integral is equivalent to


`\displaystyle\int_0^(1/2)x\cos\pi x\,\mathrm dx=\frac1{\pi^2}\int_0^(\pi/2)y\cos y\,\mathrm dy`

Now integrate by parts, setting


`u=y\implies\mathrm du=\mathrm dy`

`\mathrm dv=\cos y\implies v=\sin y`

So you have


`\displaystyle\frac1{\pi^2}\int_0^(\pi/2)y\cos y\,\mathrm dy=\frac1{\pi^2}y\sin y\bigg|_(y=0)^(y=\pi/2)-\frac1{\pi^2}\int_0^(\pi/2)\sin y\,\mathrm dy`

`=\displaystyle(\frac\pi2\sin\frac\pi2-0)/(\pi^2)+\frac1{\pi^2}\cos y\bigg|_(y=0)^(y=\pi/2)`

`=\displaystyle\frac1{2\pi}+(\cos\frac\pi2-\cos0)/(\pi^2)`

`=\displaystyle\frac1{2\pi}-\frac1{\pi^2}`