Student Question

`int sqrt(1-x^2)/x^4 dx` Find the indefinite integral

Expert Answers

An illustration of the letter 'A' in a speech bubbles

Given ,

`int sqrt(1-x^2)/x^4 dx`

By applying Integration by parts we can solve the given integral


let `u= sqrt(1-x^2)` `,v' = (1/x^4) `

=>` u' = (sqrt(1-x^2) )'`

=> =`d/dx (sqrt(1-x^2)) `

let `t=1-x^2 `


`d/dx (sqrt(1-x^2))`

=`d/dx (sqrt(t))`

= `d/(dt) sqrt(t) * d/dx (t)`        [as `d/dx f(t) = d/(dt) f(t) (dt)/dx` ]

=  `[(1/2)t^((1/2)-1) ]*(d/dx (1-x^2))`

=  `[(1/2)t^(-1/2)]*(-2x)`

=`[1/(2sqrt(1-x^2 ))]*(-2x)`


so,  `u' = -x/sqrt(1-x^2)` and as ` v'=(1/x^4)` so

`v = int 1/x^4 dx `

= `int x^(-4) dx`

= `(x^(-4+1))/(-4+1) `


= `-(1/(3x^3))`

so , let us see the values altogether.

`u= sqrt(1-x^2) ,u' = -x/sqrt(1-x^2)` and` v' = (1/x^4) ,v=-(1/(3x^3))`

Now ,by applying the integration by parts `int uv' ` is given as

 `int uv' = uv - int u'v`


`int sqrt(1-x^2)/x^4 dx `

= `(sqrt(1-x^2)) (-(1/(3x^3))) - int (-x/sqrt(1-x^2))(-(1/(3x^3))) dx `

= `(sqrt(1-x^2)) (-(1/(3x^3))) -(- int (-x/sqrt(1-x^2))((1/(3x^3))) dx) `

=` (sqrt(1-x^2)) (-(1/(3x^3))) - int (x/sqrt(1-x^2))((1/(3x^3))) dx `

= `-(sqrt(1-x^2)) ((1/(3x^3))) - int (x/sqrt(1-x^2))((1/(3x^3))) dx-----(1)`

Now let us solve ,

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

=`int (1/sqrt(1-x^2))((1/(3x^2))) dx`

=`(1/3)int (1/sqrt(1-x^2))((1/(x^2))) dx`

=`(1/3)int (1/((x^2)sqrt(1-x^2))) dx`

This integral can be solve by using the Trigonometric substitution(Trig substitution)

when the integrals containing `sqrt(a-bx^2)`then we have to take `x=sqrt(a/b) sin(t)`to solve the integral easily

so here , For

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

`x` is given as

`x= sqrt(1/1) sin(t) = sin(t)`

as `x= sin(t) `

`=>`  `dx= cos(t) dt`

now substituting the value of `x` in `(2)` we get

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

=`(1/3)int (1/(((sin(t))^2)sqrt(1-(sin(t))^2))) (cos(t) dt)`

= `(1/3)int (1/(((sin(t))^2)sqrt(cos(t))^2))) (cos(t) dt)`

=`(1/3)int (1/(((sin(t))^2)*(cos(t)))) (cos(t) dt)`

=`(1/3)int 1/(((sin(t))^2)) dt`

=`(1/3)int (csc(t))^2 dt `     

= `(-1/3) cot(t) +c`

=` (-1/3) cot(arcsin(x)) +c ---(3)`   

[since` x= sin(t) => `     `t= arcsin(x)`]

Now substituting (3) in (1) we get

(1) =>

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

=`-(sqrt(1-x^2)) ((1/(3x^3))) - ((-1/3) cot(arcsin(x)) +c)`

=`-(sqrt(1-x^2)) ((1/(3x^3)))+(1/3) cot(arcsin(x)) +c`

=`(1/3) cot(arcsin(x))- (((sqrt(1-x^2))/(3x^3))) +c----(4)`

`cot(t)` in terms of `sin(t)` can be given as follows

`cot(t) = cos(t)/sin(t) = sqrt(1-(sin(t))^2)/sin(t)`


`cot(arcsin(x)) = sqrt(1-(sin(arcsin(x)))^2)/sin(arcsin(x)) = sqrt(1-x^2)/x`

substituting the above in `(4)` we get

`(1/3) cot(arcsin(x))- (((sqrt(1-x^2))/(3x^3))) +c`

=`(1/3) (sqrt(1-x^2)/x)- (((sqrt(1-x^2))/(3x^3))) +c`

=`(sqrt(1-x^2)/(3x))- (((sqrt(1-x^2))/(3x^3))) +c`


`int sqrt(1-x^2)/x^4 dx`

=`(sqrt(1-x^2)/(3x))- sqrt(1-x^2)/(3x^3)+c`

See eNotes Ad-Free

Start your 48-hour free trial to get access to more than 30,000 additional guides and more than 350,000 Homework Help questions answered by our experts.

Get 48 Hours Free Access
Approved by eNotes Editorial