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

Question

My attempt (formula for partial integration: $\int fg = Fg - \int Fg'$):

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

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

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

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

Answer 1

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

Answer 2

HINT: for the indefinite integral we get: $$4\,\cos \left( x \right) +4\,\sin \left( x \right) x-\sin \left( x \right) \pi $$