How do you find the following indefinite integral of #(x^(2)+sin(2x))dx#? Calculus Introduction to Integration Integrals of Trigonometric Functions 1 Answer GiĆ³ Aug 26, 2015 I found: #x^3/3+sin^2(x)+c# Explanation: Consider: #int(x^2+sin(2x))dx=# #intx^2dx+intsin(2x)dx=# #=x^3/3+int2sin(x)cos(x)dx=# but #d[sin(x)]=cos(x)dx#; #x^3/3+2intsin(x)d[sin(x)]=x^3/3+cancel(2)sin^2(x)/cancel(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 1354 views around the world You can reuse this answer Creative Commons License