What is the antiderivative of #sin^2(x)#? Calculus Introduction to Integration Integrals of Trigonometric Functions 1 Answer Euan S. Jul 4, 2016 #= 1/2[x - 1/2sin2x] + C# Explanation: We're going to use the trig identity #cos2theta = 1 -2sin^2theta# #implies sin^2x = 1/2(1 - cos2x)# So #int sin^2xdx = 1/2int(1-cos2x)dx# #= 1/2[x - 1/2sin2x] + 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 95011 views around the world You can reuse this answer Creative Commons License