Question

Antiderivative of a square root

`\sqrt{1 + t ^(2) }`
​

Answer 1

Substitute
`t=\tan s`, so that
`\mathrm dt=\sec^2s\,\mathrm ds`. The integral is then equivalent to

`\displaystyle\int√(1+(\tan s)^2)\,\sec^2s\,\mathrm ds`

`=\displaystyle\int√(1+\tan^2s)\,\sec^2s\,\mathrm ds`

`=\displaystyle\int√(\sec^2s)\,\sec^2s\,\mathrm ds`

In general,
`√(x^2)=|x|`, so
`√(\sec^2s)=|\sec s|`.

We want the substitution made above to be reversible, so that
`s=\tan^(-1)t`. This restricts
`s` to the interval
`-\frac\pi2<s<\frac\pi2`, and over this interval we have
`\cos s>0\implies\sec s>0`, so we take the positive square root in order that
`√(\sec^2s)=+\sec s`.

Then the integral becomes

`=\displaystyle\int\sec^3s\,\mathrm ds`

which can be computed in several ways. One method is to integrate by parts, taking

`u=\sec s\implies\mathrm du=\sec s\tan s\,\mathrm ds`

`\mathrm dv=\sec^2s\,\mathrm ds\implies v=\tan s`

so that

`\displaystyle\int\sec^3s\,\mathrm ds=\sec s\tan s-\int\sec s\tan^2s\,\mathrm ds`

`\displaystyle\int\sec^3s\,\mathrm ds=\sec s\tan s-\int\sec s(\sec^2s-1)\,\mathrm ds`

`\displaystyle\int\sec^3s\,\mathrm ds=\sec s\tan s-\int(\sec^3s-\sec s)\,\mathrm ds`

`\displaystyle2\int\sec^3s\,\mathrm ds=\sec s\tan s+\int\sec s\,\mathrm ds`

`\displaystyle\int\sec^3s\,\mathrm ds=\frac12\sec s\tan s+\frac12\ln|\sec s+\tan s|+C`

Then with
`s=\tan^(-1)t`, you have
`\tan s=t` and
`\sec s=√(1+t^2)`, which follows from the Pythagorean identity. So

`\displaystyle\int√(1+t^2)\,\mathrm dt=\fract+√(1+t^2)2+C`