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Question: \(\int _ { 0 } ^ { \pi / 2 } \{ x - [ \sin x ] \} d x\) is equal to...

∫0π/2{x−[sin⁡x]}dx\int _ { 0 } ^ { \pi / 2 } \{ x - [ \sin x ] \} d x is equal to

A

π28\frac { \pi ^ { 2 } } { 8 }

B

π28−1\frac { \pi ^ { 2 } } { 8 } - 1

C

π28−2\frac { \pi ^ { 2 } } { 8 } - 2

D

None of these

Answer

π28\frac { \pi ^ { 2 } } { 8 }

Explanation

Solution

∫0π/2{x−[sin⁡x]}dx=∫0π/2xdx−∫0π/2[sin⁡x]dx\int _ { 0 } ^ { \pi / 2 } \{ x - [ \sin x ] \} d x = \int _ { 0 } ^ { \pi / 2 } x d x - \int _ { 0 } ^ { \pi / 2 } [ \sin x ] d x

=π28= \frac { \pi ^ { 2 } } { 8 }, [∵∫0π/2[sin⁡x]dx=0]\left[ \because \int _ { 0 } ^ { \pi / 2 } [ \sin x ] d x = 0 \right] .