`int(1+cosx)/(sinxcosx)dx`

int(cosx)/(sinx+cosx)dx=

int (1+cosx)/(x+sin x)dx

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

Evaluate the following : int_(0)^(pi//2)(sinx-cosx)/(1+sinx.cosx)dx

int(1)/(1+sinx+cosx)dx=

int(1)/(1+sinx+cosx)dx=

int(logtanx)/(sinxcosx)dx=?

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

int(dx)/(1+2sinx+cosx)