Show that `int_0^pixf(sinx)dx=pi/2int_0^pif(sinx)dxdot`

Show that int_(0)^(2pi) g(cosx)dx=2int_(0)^pi g(cosx)dx , wher g(cosx) is a function of cosx .

If f(x+f(y))=f(x)+yAAx ,y in Ra n df(0)=1, then prove that int_0^2f(2-x)dx=2int_0^1f(x)dxdot

Show that int_(0)^(pi) g (sinx)dx=2 int_(0)^((pi)/(2))g(sin)dx , where g(sinx) is a function of sinx .

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

int_0^1(tan^(-1)x)/x dx is equals to int_0^(pi/2)(sinx)/x dx (b) int_0^(pi/2)x/(sinx)dx 1/2int_0^(pi/2)(sinx)/x dx (d) 1/2int_0^(pi/2)(""x)/(sinx)dx

STATEMENT 1: int_0^pixsinxcos^2xdx=pi/2int_0^pisinxcos^2x dxdot STATEMENT 2: int_a^b xf(x)dx=(a+b)/2int_a^bf(x)dxdot

int_(0)^(pi)(cosx)/(1+sinx)dx=

int(1)/(1+sinx)dx :