I'm trying to solve this Riemann integral$\int^{\frac {\pi}{2}}_0 x \,d\sin(x) $ using the theorem* but I don't know if apply the theorem in the correct form.
We have that $F(x)=\int^{\frac {\pi}{2}}_0 x \,d\sin(x)$. Then $F'(x)=\frac {\pi}{2} (\sin x)'=\frac{\pi}{2} (\cos x)$.
Is my answer correct? Is there another way to calculate the integral?
Theorem*: Let $\alpha\in BV([a,b])$ and $f\in R(\alpha)$ in $[a,b]$ and $f$ continuous in x$\in (a,b)$. Then
a) If $\alpha$ is continuous in $c$, then $F$ is defined by F(x)= $\int^x_a f(t) \, d\alpha(t)$ is continuous in $c.
b) $F\in BV([a,b])$
c) if $\alpha$ is differentiable over $(a,b)$ then $F'(x)=f(x)\alpha'(x)$.
NOTE: Here BV means bounded variation and $R(\alpha)$ means riemann integrable over function $\alpha$
You want to solve $$ \int_0^{\frac{\pi}{2}}x\mathrm d\sin x=\int_0^{\frac{\pi}{2}}x\cos x\mathrm dx $$ which can be done by parts to find $$\int_0^{\frac{\pi}{2}}x\mathrm d\sin x=\frac{\pi}{2}-1$$