Use partial Integration to show $int_0^{frac{pi}{2}} (4x-pi)cdot cos(x) = pi – 4cdot(sqrt2-1)$

My attempt (formula for partial integration: $int fg = Fg – int Fg’$):

$F(x) = sin(x), f(x) = cos(x), g(x) = 4x- pi, g'(x) = 4$

$sin(x)(4x-pi)- int sin(x)cdot 4 = sin(x)(4x-pi)+ 4cos(x)+C$

$$int_0^{frac{pi}{2}} (4x-pi)cdot cos(x) = $$
$$[sin(x)(4x-pi)+ 4cos(x)]_0^{frac{pi}{2}} = $$
$$[sin({frac{pi}{2}})(4frac{pi}{2}-pi
)+4 cos(frac{pi}{2})]-[sin(0)(4cdot 0-pi
)+4 cos(0)] = $$
$$1cdot (2pi-pi)+4cdot 0 – 4 = $$
$$pi- 4 $$

I don’t get why $(sqrt{2}-1)$ is missing

Answer

The equality you state at the beginning is wrong. This integral is indeed $pi-4$.