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

Given ,

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

By applying Integration by parts we can solve the given integral

so,

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 `

so,

`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)`

=`-x/sqrt(1-x^2)`

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) `

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

= `-(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`

then,

`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)`

so,

`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`

so,

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

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