`int_0^(2pi)x/(1+tan^(24)x)dx , n >0` (i) `pi^2` (ii)`(pi/2)^2` (iii) `int_0^(a)f(x)dx=int_0^(a)f(a-x)dx `can be used to evaluate it (iv) Integrand is odd function

int_(0)^(2 pi)(x)/(1+tan^(24)x)dx,n>0( i) pi^(2)( ii) ((pi)/(2))^(2) (iii) int_(0)^(a)f(x)dx=int_(0)^(a)f(a-x)dx can be used to evaluate it (iv) Integrand is odd function

int_0^(pi/2)1/(1+tan^2x)dx

int_0^(pi/8) tan^(2) (2x) dx =

If int_0^(a) f(x) dx = int_0^(a) f(a-x) dx , then the value of int_0^(pi) x. f(sin x) dx =

Prove that int_(0)^(a)f(x)dx=int_(0)^(a)f(a-x)dx and hence evaluate int_(0)^(pi//2)(2log sin x-log sin2x)dx .

Prove that int_0^a f(x)dx=int_0^af(a-x)dx , hence evaluate int_0^pi(x sin x)/(1+cos^2 x)dx

If int_(0)^(pi) x f(sinx) dx=A int_(0)^((pi)/(2))f(sinx)dx , then the value of A is -