1.∫0π/2sin(x−[x]) dx=\int_0^{\pi/2} \sin(x-[x])\,dx =∫0π/2sin(x−[x])dx=, where [⋅][\cdot][⋅] denotes GIFa.3(1−cos1)+sin2−sin13(1-\cos1)+\sin2-\sin13(1−cos1)+sin2−sin1b.cos2−sin2\cos2-\sin2cos2−sin2c.3(1−cos1)+cos2−sin13(1-\cos1)+\cos2-\sin13(1−cos1)+cos2−sin1d.000Login to continueOnly logged in users canattempt or see the solution.
