`T h ev a l u eofint_0^(pi/2)sin|2x-alpha|dx ,w h e r ealpha in [0,pi],i s` `1-cosalpha` (b) `1+cosalpha` `1` (d) `cosalpha`

Answer

C

Explanation

`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`