If `f(theta) = int_0 ^oo (tan^-1x) / (x^2 + 2x cos theta + 1) dx, theta in (0 , pi) ` then `f(pi/2) = pi^2 / a` where, `a` is

f(theta)=int_0^1 (x+sin theta)^2 dx

Let f(theta) =int_(0)^(1) (x+ sin theta )^(2) dx and g(theta)=int_(0)^(1)(x+ cos theta )^(2) dx where theta in [0,2 pi] . The quantity f (theta) -g(theta), AA theta in the interval given in column I, is

Let f(theta) =int_(0)^(1) (x+ sin theta )^(2) dx and g(theta)=int_(0)^(1)(x+ cos theta )^(2) dx where theta in [0,2 pi] . The quantity f (theta) -g(theta), AA theta in the interval given in column I, is

Let f(theta) =int_(0)^(1) (x+ sin theta )^(2) dx and g(theta)=int_(0)^(1)(x+ cos theta )^(2) dx where theta in [0,2 pi] . The quantity f (theta) -g(theta), AA theta in the interval given in column I, is

Let f(x) = int_(x)^(x+(pi)/(3))|sin theta|d theta(x in [0,pi])

Evaluate : int_0^(pi//2)(dx)/(1+costhetadotcos x)theta in (0,pi)

If f(x)=|{:( sin^2 theta , cos^(2) theta , x),(cos^(2) theta , x , sin^(2) theta ),( x , sin^(2) theta , cos^2 theta ):}|, theta in (0,pi//2), then roots of f(x)=0 are

If f(theta) = sin(tan^(-1)(("sin theta)/(sqrt(cos 2 theta)))) , where -pi/4 lt theta lt pi/4 , then the value of d/(d(tan theta)) f(theta) is