How do you integrate #(sinx+secx)/tanx #? Calculus Introduction to Integration Integrals of Trigonometric Functions 1 Answer Marko T. Mar 15, 2018 Rewrite #int((sinx+1/cosx)/(sinx/cosx))dx# #=int(((sinxcosx+1)/cancelcosx)/(sinx/cancelcosx))dx# #=int(((cancelsinxcosx)/cancelsinx)+1/sinx)dx# #=int(cosxdx)+int(cscxdx)# Now both of these are standard integrals #=sinx-ln(csc(x)+cot(x)) +#Constant 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 4876 views around the world You can reuse this answer Creative Commons License