lf `alpha in (2,3)` then number ofsolution ofthe equation) `int_0^oo cos(x+alpha^2)dx=sinalpha` is .

Find int (cos2x-cos2alpha)/(six-sinalpha)dx

The value of alpha in [0,2pi] which does not satify the equation int_(pi//2)^(alpha) sin x dx = sin 2 alpha, is

The value of alpha in [0,2pi] which does not satify the equation int_(pi//2)^(alpha) sin x dx = sin 2 alpha, is

int(cos2x-cos2 alpha)/(sin x-sin alpha)dx=

int secx/(sqrt(sin(2x+alpha)+sinalpha))dx

The value of the integral int_0^(3alpha) cosec(x-alpha) cosec(x-2alpha)dx is

The value of the integral int_(0)^(3alpha) cosec (x-alpha)cosec(x-2alpha)dx is

The value of the integral int_(0)^(3alpha) cosec (x-alpha)cosec(x-2alpha)dx is

int(cos alpha)/(sin x cos(x-alpha))dx=

Let I_(1)=int_(0)^(alpha)(1+2cosx)/(1+e^(x))dx and I_(2)=int_(0)^(alpha)(1+e^(x))/(1+2cosx)dx , where alpha is the root of the equation 2 cos x - e^(x)=0 . and alpha is positive Then,