How do you find the integral of #int sec^2(5x)dx#? Calculus Introduction to Integration Integrals of Trigonometric Functions 1 Answer bp Oct 11, 2015 #1/5 tan 5x +C# Explanation: Let 5x = u , so that #dx= 1/5 du# #int sec^2 5x dx = 1/5intsec^2 u du# =#1/5 tanu +C# = #1/5 tan 5x +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 35587 views around the world You can reuse this answer Creative Commons License