What is the antiderivative of #x(sinx)^2#? Calculus Introduction to Integration Integrals of Trigonometric Functions 1 Answer MetaPhysik Jan 4, 2018 #1/4x^2-1/4xsin2x+1/8cos2x# Explanation: #I = intxsin^2xdx# #cos2x= cos^2-sin^2x , rArr sin^2x= (1-cos2x)/2 # #I = intxsin^2xdx = int(x(1-cos2x))/2dx# #I = 1/2intxdx- 1/2intxcos2xdx# Let evaluate the second part first. Let #u = 2x rArr du = 2dx# and perform integration by parts #1/2intxcos2xdx = 1/8 intucosudu= 1/8 (usinu-intsinudu)# #=1/8 (usinu+cosu)# #I = 1/4x^2-1/8(2xsin2x+cos2x)# #I= 1/4x^2-1/4xsin2x+1/8cos2x# 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 2355 views around the world You can reuse this answer Creative Commons License