Question

Evaluate the integral `int x^2/(1-x^2)^(1/2) dx`?

Answer 1

`1/2{arc sinx-xsqrt(1-x^2)}+C.`

Explanation:

Method I : Using Trigonometris Substitution.

Let, `I=intx^2/(1-x^2)^(1/2)dx.`

Sub.ing `x=siny rArr dx=cosydy.`

`:. I=intsin^2y/(1-sin^2y)^(1/2) cosydy,`

`=intsin^2y/cosy cosydy,`

`=intsin^2ydy,`

`=int(1-cos2y)/2dy,`

`=1/2{y-(sin2y)/2},`

`=1/2{y-1/2*2sinycosy},`

`=1/2{y-sinysqrt(1-sin^2y)}.`

Since, `x=siny, i.e., y=arc sinx,`

`I=1/2{arc sinx-xsqrt(1-x^2)}+C.`

Answer 2

`int \ x^2/(1-x^2)^(1/2) \ dx =1/2arcsin(x) - 1/2 xsqrt(1-x^2) + C`

Explanation:

Se week:

`I = int \ x^2/(1-x^2)^(1/2) \ dx`

Which we can write as:

`I = int \ (1+x^2-1)/sqrt(1-x^2) \ dx`
`\ \ = int \ (1-(1-x^2))/sqrt(1-x^2) \ dx`
`\ \ = int \ 1/sqrt(1-x^2) - sqrt(1-x^2) \ dx`

The first integrand is a standard integral. so we have:

`\ \ = arcsin(x) - int \ sqrt(1-x^2) \ dx`

For the second integral, consider:

`J = int \ sqrt(1-x^2)`

and perform a substitution, Let:

`x = sin theta => (dx)/(d theta)=cos theta`

Substituting we get:

`J = int \ sqrt(1-sin^2 theta) cos theta \ d theta`
`\ \ = int \ cos^2 theta \ d theta`
`\ \ = int \ 1/2(cos2 theta -1) \ d theta`
`\ \ = 1/2( (sin 2theta)/2+theta)`

And restoring the substitution we find:

`J = 1/2( (2xsqrt(1-x^2))/2 + arcsinx)`
`\ \ = 1/2( xsqrt(1-x^2) + arcsinx)`

Using this result in conjunction with the earlier result we then have:

`I = arcsin(x) - 1/2( xsqrt(1-x^2) + arcsinx) + C`
`\ \ = 1/2arcsin(x) - 1/2 xsqrt(1-x^2) + C`

Answer 3

`-x/2sqrt(1-x^2)+1/2arc sinx+C,`

Explanation:

Method II : Without Substitution.

We have, `I=intx^2/(1-x^2)^(1/2) dx,`

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

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

`=-int{(1-x^2)/sqrt(1-x^2)-1/sqrt(1-x^2)}dx,`

`=-intsqrt(1-x^2)dx+int1/sqrt(1-x^2)dx,`

`-{x/2sqrt(1-x^2)+1/2arc sinx}+arc sinx,`

`=-x/2sqrt(1-x^2)-1/2arc sinx+arc sinx,`

`=-x/2sqrt(1-x^2)+1/2arc sinx+C,` as in Method I !