`IfA=int_0^pi(cosx)/((x+2)^2)dx ,t h e nint_0^(pi/2)(sin2x)/(x+1)dxi se q u a lto` `1/2+1/(pi+2)-A` (b) `1/(pi+2)-A` `1+1/(pi+2)-A` (d) `A-1/2-1/(pi+2)`

Options

A. `1/2+1/(pi+2)-A`
B. `1/(pi+2)-A`
C. `1+1/(pi+2)-A`
D. `A-1/2-1/(pi+2)`

Answer

A

Explanation

`I=int_(0)^(pi//2) (sin 2x)/(x+1)dx`. Put `x=y//2`. Then
`I=int_(0)^(pi)(siny)/(y+2)dy`
`=((-cosy)/(y+2))_(2)^(pi)-int_(0)^(pi) (cosy)/((y+2)^(2))dy` (Integrating by parts)
`1/(pi+2)+1/2-A`