Question

If we let x=tan theta, then integral of (1+x^2)^(1/2) dx from 1to 3^(1/2) is equivalent to

Answer 1

`\displaystyle\int_(x=1)^(x=\sqrt3)√(1+x^2)\,\mathrm dx`

With
`x=\tan t`, we have
`\mathrm dx=\sec^2t\,\mathrm dt` and
`t=\arctan x`. So the integral is equivalent to


`\displaystyle\int_(t=\arctan1)^(t=\arctan\sqrt3)√(1+\tan^2t)\sec^2t\,\mathrm dt`

`=\displaystyle\int_(t=\pi/4)^(t=\pi/3)√(\sec^2t)\sec^2t\,\mathrm dt`

`=\displaystyle\int_(t=\pi/4)^(t=\pi/3)\sec^3t\,\mathrm dt`

which is a fairly standard integral to compute. Using the power reduction formula or integrating by parts, you find


`\displaystyle=\frac12\sec t\tan t\bigg|_(t=\pi/4)^(t=\pi/3)+\frac12\int_(t=\pi/4)^(t=\pi/3)\sec t\,\mathrm dt`

`\displaystyle=\frac{2\sqrt3-\sqrt2}2+\frac12\ln|\sec t+\tan t|\bigg|_(t=\pi/4)^(t=\pi/3)`

`=\frac{2\sqrt3-\sqrt2}2+\frac12(\ln(2+\sqrt3)-\ln(1+\sqrt2))`