How do you find the antiderivative of #(sinx + cosx)/tanx#? Calculus Introduction to Integration Integrals of Trigonometric Functions 1 Answer Ratnaker Mehta Jul 27, 2016 #sinx+ln|tan(x/2)|+cosx+C# Explanation: # I=int(sinx+cosx)/tanxdx=int(sinx+cosx)cosx/sinxdx# #=intcosxdx+intcos^2x/sinxdx=sinx+int(1-sin^2x)/sinxdx# #=sinx+int{1/sinx-sinx}dx# #=sinx+int(cscx-sinx)dx# #:. I=sinx+ln|tan(x/2)|+cosx+C# 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 1440 views around the world You can reuse this answer Creative Commons License