How do you integrate #((cosx)^3 sinx )/(4sqrt{1-(cosx)^4})dx#?

Question

2408 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-04-17T19:01:22

This integral is very easy, explanation :

#1/4int(cos^3(x)*sin(x))/(sqrt(1-cos^4(x)))dx #

Let's #t=cos(x)#

So #dt = -sin(x)dx#

#1/2intt^3/(2(sqrt(1-t^4)))d(-t) = 1/2int(-t^3)/(2sqrt(1-t^4))dt #

Let's #u = 1-t^4#

So #du = -4t^3dt#

#=>1/8int(-4t^3)/(2sqrt(1-t^4))dt#

#=>1/8int1/(2sqrt(u))du#

#= 1/8[sqrt(u)]#

Substitute back for #u = 1-t^4# and #t=cos(x)#

#= 1/8[sqrt(1-cos^4(x))] + C#

With habits you can directly do #t = 1 - cos^4(x)#

#dt = 4sin(x)cos^3(x)#