What's the integral of #int 1/(secx+tanx)dx#? Calculus Introduction to Integration Integrals of Trigonometric Functions 1 Answer ali ergin Mar 6, 2016 #int 1/(sec x+tan x)d x=l n(1+sin x) + C# Explanation: #int 1/(sec x+tan x)d x=(d x)/(1/cos x+sin x/cos x)# #int (cos x d x)/(1+sin x)" "1+sin x=u" "cos x d x=d u# #int (d u)/u# #int 1/(sec x+tan x)d x=l n u +C# #int 1/(sec x+tan x)d x=l n(1+sin x) + 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 17160 views around the world You can reuse this answer Creative Commons License