assuming it's #int 1/(sinx+cosx) dx#
#sinx+cosx = sinx+sin(x+pi/2)#
You remember this unknown formula ? :
#sin(q)+sin(p) = 2sin((q+p)/2)cos((q-p)/2)#
In this case we have :
#sinx+sin(x+pi/2) = 2sin(x+pi/4)cos(pi/4) = sqrt(2)sin(x+pi/4)#
#int 1/(sinx+cosx) dx = 1/sqrt(2)int1/sin(x+pi/4)dx#
#t = x+pi/4#
#dt = dx#
#1/sqrt(2)int1/sin(t)dt#
#-1/sqrt(2)int-sin(t)/sin^2(t)dt#
#u = cos(t)#
#du = -sin(t)dt#
With pythagore #sin^2t = 1-cos^2t#
#-1/sqrt(2)int1/(1-u^2)dt#
Which is the derivate of #-1/sqrt2arctanh(u)#
#[-1/sqrt2arctanh(u)]#
Substitute back
#[-1/sqrt2arctanh(cos(t))]#
#[-1/sqrt2arctanh(cos(x+pi/4))]#
