What is the integral of #int [4sec(x)]/[cos(x)] dx#? Calculus Introduction to Integration Integrals of Trigonometric Functions 1 Answer Konstantinos Michailidis Mar 20, 2016 It is #int [4sec(x)]/[cos(x)] dx=int 4/(cos^2 x)dx=4*tanx+c# Note #secx=1/cosx# 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 1203 views around the world You can reuse this answer Creative Commons License