`I_(1)=int_(0)^((pi)/2)(sinx-cosx)/(1+sinxcosx)dx, I_(2)=int_(0)^(2pi)cos^(6)dx`,
`I_(3)=int_(-(pi)/2)^((pi)/2)sin^(3)xdx, I_(4)=int_(0)^(1) In (1/x-1)dx`. Then

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

int_(0)^((pi)/(2))(cosx-sinx)/(1+sinxcosx)dx

int_(0)^((pi)/(2))(cosx-sinx)/(1+sinxcosx)dx

int_(0)^(pi//2)((sinx-cosx))/((1+sinxcosx))dx=0

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

I_1=int_0^(pi/2)(sinx-cosx)/(1+sinxcosx)dx ,I_2=int_0^(2pi)cos^6xdx , I_3=int_(-pi/2)^(pi/2)sin^3xdx ,I_4=int_0^1 1n(1/x-1)dxdot Then

Evaluate int_0^(pi/2) (sinx-cosx)/(1+sinxcosx)dx

If I_(1)=int_(0)^(2pi)sin^(3)xdx and I_(2)=int_(0)^(1)ln((1)/(x)-1)dx , then

If I_(1)=int_(0)^(2pi)sin^(3)xdx and I_(2)=int_(0)^(1)ln((1)/(x)-1)dx , then