What is the integral of #int tan^4(x)sec^2(x)#? Calculus Introduction to Integration Integrals of Trigonometric Functions 1 Answer Shwetank Mauria Jun 19, 2016 #inttan^4xsec^2xdx=tan^5x/5+c# Explanation: Let #u=tanx# #du=sec^2xdx# Hence, #inttan^4xsec^2xdx# = #intu^4du# = #u^5/5+c# = #tan^5x/5+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 26091 views around the world You can reuse this answer Creative Commons License