How do I evaluate the indefinite integral #intcot^5(x)*sin^4(x)dx# ?

1 Answer
Apr 23, 2018

Answer:

#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"#