What is the integral of #cos^(2)2xdx#? Calculus Introduction to Integration Integrals of Trigonometric Functions 1 Answer Narad T. Dec 9, 2016 The answer is #=x/2+(sin4x)/8+C# Explanation: We use #cos2theta=2cos^2theta-1# #cos^2theta=(1+cos2theta)/2# Here #theta=2x# So, #cos^2(2x)=(1+cos4x)/2# #intcos^2(2x)dx=int((1+cos4x)dx)/2# #=1/2(x+sin(4x)/4)+C# #=x/2+(sin4x)/8+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 1278 views around the world You can reuse this answer Creative Commons License