Question

`int_(-a)^adx/((a^2+x^2)sqrt((2a^2+x^2)))=?`

Answer 1

`int_(-a)^(a) \ 1/((a^2+x^2)sqrt(2a^2+x^2)) \ dx = pi/(3a^2)`

Explanation:

We seek the definite integral:

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

For simplicity, consider the indefinite integral:

`J = int \ 1/((a^2+x^2)sqrt(2a^2+x^2)) \ dx`

Let us perform a substitution, where we define:

`x = sqrt(2)a tanu => dx/(du) = sqrt(2)a sec^2 u`

Then Substituting we get:

`J = int \ (sqrt(2)a sec^2 u)/((a^2+2a^2tan^2u)sqrt(2a^2+2a^2tan^2u)) \ du`

`\ \ = int \ (sqrt(2)a sec^2 u)/(a^2(1+2tan^2u)sqrt(2a^2(1+tan^2u))) \ du`

`\ \ = int \ (sqrt(2)a sec^2 u)/(a^2(1+2tan^2u) \ sqrt(2)a sqrt(1+tan^2u)) \ du`

`\ \ = 1/a^2 \ int \ (sec u)/(1+2tan^2u) \ du`

`\ \ = 1/a^2 \ int \ (sec u)/(1+2(sec^2u-1)) \ du`

`\ \ = 1/a^2 \ int \ (sec u)/(2sec^2u-1) \ du`

`\ \ = 1/a^2 \ int \ (1/(cos u))/(2/cos^2u-1) \ du`

`\ \ = 1/a^2 \ int \ (1/(cos u))/((2-cos^2u)/cos^2u) \ du`

`\ \ = 1/a^2 \ int \ (cos u)/(2-cos^2u) \ du`

`\ \ = 1/a^2 \ int \ (cos u)/(2-(1-sin^2u)) \ du`

`\ \ = 1/a^2 \ int \ (cos u)/(1+sin^2u) \ du`

And we can perform a second substitution, where we define:

`v=sinu => (dv)/(du) = cosu`

Then Substituting we get:

`J = 1/a^2 \ int \ (1)/(1+v^2) \ dv`

Which is a standard integral, and so integrating we get:

`J = 1/a^2 \ arctan(v) + C`

Then, we restore the earlier `v` substitution:

`J = 1/a^2 \ arctan(sinu) + C`

And prior to restoring the `u` substitution we use the identity

`\ \ \ \ \ 1 + tan^2u -= sec^2 u`
`:. 1 + (x/(sqrt(2)a))^2 = 1/(1-sin^2u)`
`:. 1 + x^2/(2a^2) = 1/(1-sin^2u)`
`:. (2a^2 + x^2)/(2a^2) = 1/(1-sin^2u)`
`:. 1-sin^2u = (2a^2)/(2a^2 + x^2)`
`:. sin^2u = 1-(2a^2)/(2a^2 + x^2)`
`:. sin^2u = (2a^2 + x^2-2a^2)/(2a^2 + x^2)`
`:. sin^2u = (x^2)/(2a^2 + x^2)`
`sinu = sqrt((x^2)/(2a^2 + x^2))`

Which now allows us to restore the `u` substitution:

`J = 1/a^2 \ arctan( x/(sqrt(2a^2 + x^2)) ) + C`

So now we can evaluate the definite integral:

`I = [1/a^2 \ arctan( x/(sqrt(2a^2 + x^2)) )]_(-a)^(a)`

`\ \ = 1/a^2 { arctan( a/(sqrt(2a^2 + a^2))) - arctan( (-a)/(sqrt(2a^2 + (-a)^2))) }`

`\ \ = 1/a^2 { arctan( 1/sqrt(3)) - arctan( -1/sqrt(3)) }`

`\ \ = 1/a^2 { pi/6 - (-pi/6) }`

`\ \ = 1/a^2 pi/3`

`\ \ = pi/(3a^2)`