(sqrt(5+cos x)-2)/((pi-x)^(2))

If f(x) is continuous at x = pi , where f(x)=(sqrt(2+cos x)-1)/((pi-x)^(2))", for " x!= pi , then f(pi)=

int_((pi)/(3))^((pi)/(2))(sqrt(1+cos x))/((1-cos x)^((5)/(2)))dx

(i) lim_(x to (pi)/(2)) (1+cos 2x)/((pi-2x)^(2)) (ii) lim_( x to 0) (1-cos x.sqrt(cos 2x))/(x^(2)) (iii) lim_(thetararr(pi)/(6)) (sin(theta-(pi)/(6)))/(sqrt(3)-2 cos theta)

overset((pi)/(2))underset((pi)/(3))int (sqrt(1+cosx))/((1-cos x)^((5)/(2)))dx

Evaluate lim_(x->pi)(sqrt(1-cos6x))/(sqrt2(pi/3-x))

If pi

The value of int_(0)^( pi/2)(sqrt(cos x))/(sqrt(sin x)+sqrt(cos x))dx

Domain of the function f(x)=sqrt(cos(sin x))+sin(x^(2)-1) is [-1,1][-2,2]c*[-pi,sqrt(2)]uu[sqrt(2),pi][-sqrt(2),-sqrt(2)]

cos x-sin x = sqrt (2) cos (x + (pi) / (4))