How do you find the integral of #int cos^3x sin^2x dx#? Calculus Introduction to Integration Integrals of Trigonometric Functions 1 Answer sakura · Stefan V. Sep 30, 2015 #intcos^3xsin^2xdx=(1/3)sin^3x-(1/5)*sin^5x+c# Explanation: #intcos^3xsin^2xdx# #=intcos^2x*sin^2x*(cosx)dx# #=int (1-sin^2)sin^2x*d(sinx)# Hints: #(d(sinx))/dx=cosx# #d(sinx)=cosxdx# Therefore, #=int(sin^2x-sin^4x)*d(sinx)# #==(1/3)sin^3x-(1/5)*sin^5x+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 49355 views around the world You can reuse this answer Creative Commons License