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

$\displaystyle\int_{x=1}^{x=\sqrt3}\sqrt{1+x^2}\,\mathrm dx$

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

$\displaystyle\int_{t=\arctan1}^{t=\arctan\sqrt3}\sqrt{1+\tan^2t}\sec^2t\,\mathrm dt$
$=\displaystyle\int_{t=\pi/4}^{t=\pi/3}\sqrt{\sec^2t}\sec^2t\,\mathrm dt$
$=\displaystyle\int_{t=\pi/4}^{t=\pi/3}\sec^3t\,\mathrm dt$

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

$\displaystyle=\frac12\sec t\tan t\bigg|_{t=\pi/4}^{t=\pi/3}+\frac12\int_{t=\pi/4}^{t=\pi/3}\sec t\,\mathrm dt$
$\displaystyle=\frac{2\sqrt3-\sqrt2}2+\frac12\ln|\sec t+\tan t|\bigg|_{t=\pi/4}^{t=\pi/3}$
$=\dfrac{2\sqrt3-\sqrt2}2+\frac12(\ln(2+\sqrt3)-\ln(1+\sqrt2))$