Question

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

Answer 1

`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`