$\int_0^{π/4}\sin x\, d(x-[x])$ is equal to (where [*] denotes the greatest integer function)
Answer & explanation
Correct answer: option 2
$∵ 0 ≤ x < π/4$ $∴ [x] = 0$
Then, $\int_0^{π/4}\sin x\,d(x- [x])=\int_0^{π/4}\sin x\,dx$
$=-\{\cos x\}_0^{π/4}=-(\frac{1}{\sqrt{2}}-1)=1-\frac{1}{\sqrt{2}}$