How do you integrate #(1-tanx)/(1+tanx) dx#? Calculus Introduction to Integration Integrals of Trigonometric Functions 1 Answer Roy E. Dec 5, 2016 Rewrite as sines and cosines. Explanation: #(1-tan x)/(1+tan x)=(1-(sin x)/(cos x ))/(1+(sin x)/(cos x)) = (cos x - sin x )/(cos x + sin x ) = (cos x cos (pi/4)-sin x sin (pi/4))/(cos x cos (pi/4) + sin x sin (pi/4) )=cot(x+pi/4)# Note that #sin(pi/4)=cos(pi/4)=1/sqrt 2# Answer link Related questions How do I evaluate the indefinite integral #intsin^3(x)*cos^2(x)dx# ? How do I evaluate the indefinite integral #intsin^6(x)*cos^3(x)dx# ? How do I evaluate the indefinite integral #intcos^5(x)dx# ? How do I evaluate the indefinite integral #intsin^2(2t)dt# ? How do I evaluate the indefinite integral #int(1+cos(x))^2dx# ? How do I evaluate the indefinite integral #intsec^2(x)*tan(x)dx# ? How do I evaluate the indefinite integral #intcot^5(x)*sin^4(x)dx# ? How do I evaluate the indefinite integral #inttan^2(x)dx# ? How do I evaluate the indefinite integral #int(tan^2(x)+tan^4(x))^2dx# ? How do I evaluate the indefinite integral #intx*sin(x)*tan(x)dx# ? See all questions in Integrals of Trigonometric Functions Impact of this question 33934 views around the world You can reuse this answer Creative Commons License