How do you find the Integral of #(1-x^2)^(3/2)/ (x^6)#?

Question

4943 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-10-12T15:41:13

Go !

leeeeeeet's

#x = cos(t)#
#arccos(x) = t#

#dx = -sin(t) dt#

Integral become

#-int (sin^2(t))^(3/2)/cos^6(t)*sin(t)dt#

#-int sin^4(t)/cos^6(t)#

#-int tan^4(t)* 1/cos^2(t)dt#

Let's #u = tan(t)#

#du = 1/cos^2(t) dt#

Integral become

#-intu^4du#

#-[1/5u^5]#

Substitute back

#-[1/5tan^5(t)]#

#-[1/5tan^5(arccos(x))]#

You can simplify in

#-(1 - x^2)^(5/2)/(5 x^5)#