# How does $\pi$ become $\pi^2$?

I am currently doing integration and cannot seem to understand this question

$x\cos\left({\pi}x\right)$

I have tried following the following

Formula I have used:$∫f g′= f g − ∫ f′g$
f = $x$
f'= $1$
g = $\dfrac{\sin\left({\pi}x\right)}{{\pi}}$
g'= $cos({\pi}x)$

The end result should be the following:

$\dfrac{{\pi}x\sin\left({\pi}x\right)+\cos\left({\pi}x\right)}{{\pi}^2}$

I cannot seem to understand how that ends up being the end result.

• Which part exactly do you not understand? – MisterRiemann Jul 20 '18 at 13:37
• Did you integrate $f'g$? If you did, you get $\pi^2$ in the denominator. The rest is just finding a common denominator. – Dan Sp. Jul 20 '18 at 13:43

$$\int udv = uv - \int vdu$$
$$u = x$$ $$du = dx$$ $$dv = \cos(\pi x)$$ $$v = \frac{\sin(\pi x)}{\pi}$$
$$=\frac{x\sin(\pi x)}{\pi} - \int \frac{\sin(\pi x)}{\pi}dx$$ $$=\frac{x\sin(\pi x)}{\pi} + \frac{\cos(\pi x)}{\pi^2} + C$$ $$= \frac{\pi x \sin(\pi x) + \cos(\pi x)}{\pi^2} +C$$
As you said: $$\int x \cos(x) dx = x\frac{sin(\pi x)}{\pi} - \int \frac{sin(\pi x)}{\pi}$$
So $\pi^2$ origin from the last integral. Infact: $$\int \frac{sin(\pi x)}{\pi} = - \frac{cos^2(\pi x)}{\pi^2}$$