How do I evaluate the indefinite integral #intcot^5(x)*sin^4(x)dx# ? Calculus Introduction to Integration Integrals of Trigonometric Functions 1 Answer Monzur R. Apr 23, 2018 #cos^2x+1/4sin^4x+lnabssinx+"c"# Explanation: #cot^5xsin^4x=cos^5x/sin^5xsin^4x=cos^5x/sinx=cos^4xcotx=(1-sin^2x)^2cotx=(1-2sin^2x+sin^4x)cot=cotx-2sinxcosx+sin^3xcosx# So #intcot^5xsin^4xdx=intcotxdx-int2sinxcosxdx+intsin^3xcosxdx# Now let #u=sinx# and #du=cosxdx# and #v=cosx# and #dv=-sinxdx# #intcotxdx+int-2sinxcosxdx+intsin^3xcosxdx=lnabssinx int2vdv +intu^3du=lnabssinx+v^2+1/4u^4+"c"=cos^2x+1/4sin^4x+lnabssinx+"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 #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# ? How do I evaluate the indefinite integral #intsin(x)/(cos^3(x))dx# ? See all questions in Integrals of Trigonometric Functions Impact of this question 12657 views around the world You can reuse this answer Creative Commons License