Calculate $\int\limits_{0}^{\pi}(x-1)d(xsgn(\cos x))$.
I used fact that if $f$ is continuous and $g'$ is Riemann integrable over the specified interval, then:
$\int f(x)dg(x)=\int f(x)g'(x)dx.$
So for our integral, we have:
$\int\limits_{0}^{\pi}(x-1)d(xsgn(\cos x))= \int\limits_{0}^{\frac{\pi}{2}}(x-1)(x)'dx + \int\limits_{\frac{\pi}{2}}^{\pi}(x-1)(-x)'dx=\ldots.$
Is it correct way to solve this example?