How do you find the integral of #(sinx)^3 dx#? Calculus Introduction to Integration Integrals of Trigonometric Functions 1 Answer Konstantinos Michailidis Sep 25, 2015 Refer to explanation Explanation: We know that #sin 3x = 3sinx - 4 (sin x)^3=>4(sinx)^3=3sinx-sin3x=> (sinx)^3=3/4*sinx-1/4sin3x # Hence we have that #int (sinx)^3 dx=int (3/4*sinx-1/4*sin(3x))dx=-3/4cosx+1/12cos3x+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 8527 views around the world You can reuse this answer Creative Commons License