#int(1-x^2)^(3/2)/x^6dx# ?

1 Answer
Feb 7, 2018

#I=-(1-x^2)^(5/2)/(5x^5)+C#

Explanation:

We want to solve the integral

#I=int(1-x^2)^(3/2)/x^6dx#

Use substitution:

Let #x=cos(u)=>dx/(du)=-sin(u)#

#I=int(1-cos^2(u))^(3/2)/cos^6(u)*(-sin(u))du#

#=-intsin^4(u)/cos^6(u)du#

#=-inttan^4(u)sec^2(u)du#

Let #s=tan(u)=>(ds)/(du)=sec^2(u)#

#I=-ints^4ds#

#=-1/5s^5+C#

Substitute #s=tan(u)# and #u=arccos(x)#

#I=-1/5tan^5(arccos(x))+C#

Or using #tan(arccos(x))=sqrt(1-x^2)/x#

#I=-(1-x^2)^(5/2)/(5x^5)+C#