What is the antiderivative of #x sin (x/2)#? Calculus Introduction to Integration Integrals of Trigonometric Functions 1 Answer Andrea S. May 17, 2018 #int xsin(x/2)dx =-2xcos(x/2) + 4sin(x/2)+C# Explanation: Evaluate the indefinite integral: #int xsin(x/2)dx# by parts: #int xsin(x/2)dx = int x d/dx (-2cos(x/2))dx# #int xsin(x/2)dx =-2xcos(x/2) + 2 int cos(x/2)dx# #int xsin(x/2)dx =-2xcos(x/2) + 4 int cos(x/2)d(x/2)# #int xsin(x/2)dx =-2xcos(x/2) + 4sin(x/2)+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 1284 views around the world You can reuse this answer Creative Commons License