How do you find the integral of #sin^2 (x)cos^2 (x) dx#? Calculus Introduction to Integration Integrals of Trigonometric Functions 1 Answer Ratnaker Mehta Jul 19, 2016 #1/32(4x-sin4x)+C#. Explanation: Let us note that, #sin^2xcos^2x=1/4(sin2x)^2=sin^2(2x)/4# Recall that, #sin^2theta=(1-cos2theta)/2# Letting #theta=2x#, we have, #sin^2(2x)=(1-cos4x)/2# #:. intsin^2xcos^2xdx=1/8int(1-cos4x)dx=1/8(x-sin(4x)/4)=1/32(4x-sin4x)+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 1080 views around the world You can reuse this answer Creative Commons License