I=int_(0)^( pi/2)(1)/(6-cos x)dx

int_(0)^(pi//2)(1)/(2+cos x)dx=

int_(0)^(pi/2)(x)/(1+cos x)dx =

Find I=int_(0)^( pi)ln(1+cos x)dx

I=int_(0)^(2 pi)cos^(-1)(cos x)dx

int_(0)^( pi)(dx)/(1+cos x)

int_(0)^(pi//4) (1)/(1+cos 2x)dx

If int_(0)^( pi)(1)/(a+b cos x)dx=(pi)/(sqrt(a^(2)-b^(2))), then int_(0)^( pi)(1)/((a+b cos x)^(2))dx is

int_(0)^( pi/2)(sin x)/(1+cos x)dx

If A=int_((rho)/(2))^((pi)/(2))cos(sin x)dxB=int_(0)^((pi)/(2))sin(cos x)dx and C=int_(0)^((pi)/(2))cos(x)dx

I_(1)=int_(0)^((pi)/(2))(sin x-cos x)/(1+sin x cos x)dx,I_(2)=int_(0)^(2 pi)cos^(6)xdx,I_(3)=int_((pi)/(2))^((pi)/(2))sin^(3)xdx,I_(4)=int_(0)^(1)1n((1)/(x)-1)dx. Then I_(1)=I_(3)=I_(4)=0,I_(1)!=0I_(1)=I_(3)=0,I_(4)!=0I_(1)=I_(2)=0,I_(4)!=0I_(1)=I_(2)=I_(3)=0,I_(4)!=0