# 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