How do you find the integral of #int [cot^5 (x) (sin^4(x) dx]#? Calculus Introduction to Integration Integrals of Trigonometric Functions 1 Answer Sasha P. Oct 7, 2015 #I=ln|sinx|-sin^2x+sin^4x/4+C# Explanation: #int cot^5xsin^4xdx=int (cos^5x)/(sin^5x)sin^4xdx=# #int (cos^4xcosx)/(sinx)dx=int ((1-sin^2x)^2cosx)/sinx dx=I# #sinx=t => cosxdx=dt# #I=int (1-t^2)^2/tdt=int (1-2t^2+t^4)/tdt=int (1/t-2t+t^3)dt# #I=ln|t|-t^2+t^4/4+C# #I=ln|sinx|-sin^2x+sin^4x/4+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 8986 views around the world You can reuse this answer Creative Commons License