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The value of int0^(pi/2)sin|2x-alpha|dx ...

The value of `int_0^(pi/2)sin|2x-alpha|dx` , where `alpha in [0,pi]` , is

A

`1-cos alpha`

B

`1+cos alpha`

C

`1`

D

`cos alpha`

Text Solution

Verified by Experts

The correct Answer is:
C
`I=int_(0)^(pi//2) sin|2x-alpha|dx`
Put `2x-alpha=t`
`:. I=1/2int_(-alpha)^(pi-alpha) sin|t|dt`
`=1/2 int_(-alpha)^(0)-sin t dt +1/2 int_(0)^(pi-alpha) sin t dt ( :' alpha epsilon [0,pi])`
`=1/2[1-cos alpha]-1/2[-cos alpha-1]`
`=1`
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