How do you find the antiderivative of #(sinx)^2#? Calculus Introduction to Integration Integrals of Trigonometric Functions 1 Answer Eddie Jul 2, 2016 # = x/2 -1/4 sin 2x + C # Explanation: us the identity #cos 2A = 1 - 2 sin^2 A implies sin^2 A = 1/2(1-cos 2A)# so #int dx qquad sin^2 x# #= 1/2 int dx qquad 1 - cos 2x# # = 1/2 (x -1/2 sin 2x) + C # # = x/2 -1/4 sin 2x + 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 1100 views around the world You can reuse this answer Creative Commons License