Answer
$\frac{5\pi}{6}$+1-$\sqrt3$
Work Step by Step
Let's use a right triangle shown by the image below. Based on the trigonometric relationship, $sec^{-1}\sqrt{\theta}$ =$\beta$. Hence, sec$\beta$=$\sqrt\theta$ and $sec^{2}\beta$=$\theta$. And $\frac{d\theta}{d\beta}$=2$sec^{2}$$\beta$tan$\beta$
So, $\int$$sec^{-1}\sqrt{\theta}$d$\theta$= $\int$$\beta$($\frac{d\theta}{d\beta})$d$\beta$=2$\int$$\beta$$sec^{2}$$\beta$tan$\beta$d$\beta$ (We apply chain rule here)
To evaluate $\int$$\beta$$sec^{2}\beta$tan$\beta$d$\beta$, let u=$\beta$ and dv= $sec^{2}\beta$tan$\beta$d$\beta$. So, v=$\int$$sec^{2}\beta$tan$\beta$d$\beta$.
To evaluate v, we let p=tan$\beta$, so $sec^{2}\beta$$d\beta$=dp. v=$\int$p dp=$\frac{1}{2}$$p^{2}$=$\frac{1}{2}$$tan^{2}\beta$.
Now, using partial integration,
$\int$$sec^{-1}\sqrt{\theta}$d$\theta$= 2($\int$$\beta$$sec^{2}\beta$tan$\beta$d$\beta$)=2($\frac{1}{2}$$\beta$$tan^{2}\beta$-$\frac{1}{2}$$\int$$tan^{2}\beta$d$\beta$)
Here we use the identity $tan^{2}\beta$=$sec^{2}\beta$-1 to evaluate $\int$$tan^{2}\beta$d$\beta$
Therefore, the expression = $\beta$$tan^{2}\beta$-$\int$($sec^{2}\beta$-1)d$\beta$
Then we split the integral, =$\beta$$tan^{2}\beta$-$\int$$sec^{2}\beta$d$\beta$+$\int$d$\beta$
$\int$$sec^{2}\beta$d$\beta$=tan$\beta$, $\beta$$tan^{2}\beta$-tan$\beta$+$\beta$
Finally we convert $\beta$ back to $\theta$, expression=($sec^{-1}\sqrt{\theta}$)($\theta$-1)-$\sqrt{\theta-1}$+$sec^{-1}\sqrt{\theta}$
This gives: $\frac{5\pi}{6}$+1-$\sqrt3$.
