What is the integral of #(sinx)^3 dx# from 0 to #pi/2#? Calculus Introduction to Integration Integrals of Trigonometric Functions 1 Answer George C. Aug 7, 2015 #2/3# Explanation: #int_0^(pi/2) sin^3x dx = int_0^(pi/2) (1-cos^2x)sin x dx# #=int_0^(pi/2)(sin x -cos^2x sin x) dx# #=(-cos x+1/3cos^3 x )|_0^(pi/2)# #=(-0+1/3*0)-(-1+1/3)# #=2/3# 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 48449 views around the world You can reuse this answer Creative Commons License