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

Find the number points where f (theta) = int _(-1)^(1) (sin theta dx )/(1-2x cos theta + x ^(2)) is discontinous where thetain [0, 2pi].

Find the number points where f (theta) = int _(-1)^(1) (sin theta dx )/(2x cos theta + x ^(2)) is discontinous where thetain [0, 2pi].

The function f(theta)=(d)/(d(theta))int_(0)^( theta)(dx)/(1-cos theta cos x) satisfies the differential equation

int_ (0) ^ (pi / 2) ((theta) / (sin theta)) ^ (2) d theta =

int_0 2pi |sin3theta| d theta

If f(theta)=(cos i theta-i sin i theta),i=sqrt(-1) then the value of int_(-3)^(-1)[f(theta)+f(2 theta)+f(3 theta)+...oo]dx=

If int_(pi//2)^(theta) sin x dx=sin 2 theta then the of theta satisfying 0 lt theta lt pi , is

If int_(pi//2)^(theta) sin x dx=sin 2 theta then the of theta satisfying 0 lt theta lt pi , is