What is #int sec^2(5x) dx #? Calculus Introduction to Integration Integrals of Trigonometric Functions 1 Answer sente Feb 11, 2016 #intsec^2(5x)dx = 1/5tan(5x)+C# Explanation: Noting that #d/dx tan(x) = sec^2(x)# Let #u = 5x => du = 5dx => dx = 1/5du# #intsec^2(5x)dx = intsec^2(u)/5du# #=1/5intsec^2(u)du# #=1/5tan(u)+C# #=1/5tan(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 5859 views around the world You can reuse this answer Creative Commons License