What is the integral of #cos^(2)3x dx#? Calculus Introduction to Integration Integrals of Trigonometric Functions 1 Answer Lucy Apr 14, 2018 #1/2(1/6sin6x +x) +c# Explanation: #int (cos3x)^2 dx# =#int 1/2(cos6x +1) dx# =#1/2intcos 6x+1 dx# =#1/2(1/6sin6x +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 62162 views around the world You can reuse this answer Creative Commons License