If `f(x)=(sinx)/xAAx in (0,pi],` prove that `pi/2int_0^(pi/2)f(x)f(pi/2-x)dx=int_0^pif(x)dx`

Explanation

`int_(0)^(pi)f(x)dx=int_(0)^(pi)(sin x)/x dx`
Let `I=int_(0)^(pi//2)f(x)f((pi)/2-x)dx`
`=int_(0)^(pi//2)(sinx)/x ("sin"((pi)/2-x))/((pi)/2-x)dx`
`=int_(0)^(pi//2)(2sinx cosx)/(x(pi-2x)) dx`
`=int_(0)^(pi//2)(sin2x)/(x(pi-2x))dx`
Let `2x=t`. So `dt=2dx`
`:. I=int_(0)^(pi)(sint)/(t/2(pi-t)^(2))=int_(0)^(pi)(sint)/(t(pi-t))dt`
`=1/(pi) int_(0)^(pi) sin t(1/t+1/(pi-t))dt`
`=1/(pi) int_(0)^(pi)(sint)/t dt+1/(pi) int_(0)^(pi)(sint)/((pi-t))dt`
`=1/(pi) int_(0)^(pi) (sint)/t dt+1/(pi) int_(0)^(pi)(sin(pi-t))/(pi-(pi-t))dt`
`=1/(pi) int_(0)^(pi)(sint)/t dt+1/(pi) int_(0)^(pi) (sint)/t dt=2/(pi) (sint)/t dt`
or`(pi)/2 int_(0)^(pi//2)f(x)f(pi/2-x)dx=int_(0)^(pi)(sinx)/x dx`