Integration, get stuck at x=tan($\theta$) when calculating arc length I am learning to calculate the arc length by reading a textbook, and there is a question

However, I get stuck at calculating
$$\int^{\arctan{\sqrt15}}_{\arctan{\sqrt3}} \frac{\sec{(\theta)} (1+\tan^2{(\theta)})} {\tan{\theta}} d\theta$$ How can I continue calculating it?
Update 1:
$$\int^{\arctan{\sqrt{15}}}_{\arctan{\sqrt3}} \frac{\sec{(\theta)} (1+\tan^2{(\theta)})} {\tan{\theta}} d\theta = \int^{\arctan{\sqrt{15}}}_{\arctan{\sqrt3}} (\csc{(\theta)} + \sec{(\theta)} \tan{(\theta)}) d\theta \\ 
= \int^{\arctan{\sqrt{15}}}_{\arctan{\sqrt3}} \csc{(\theta) d\theta + \frac{1}{\cos{(\theta)}}} |^{arctan{\sqrt{15}}}_{arctan{\sqrt3}} \\
= \int^{\arctan{\sqrt{15}}}_{\arctan{\sqrt3}} \csc{(\theta) d\theta + \frac{1}{\cos{(\sqrt{15})}} - \frac{1}{\cos{(\sqrt3)}}}$$
But how can I get the final result?
Update 2:
Because $\frac{1}{\cos{(x)}} = \sqrt{ \frac{\cos^2{(x)} + \sin^2{(x)}}{cos^2{(x)}}} = \sqrt{1+\tan^2{(x)}}$, I get 
$$\frac{1}{\cos{(\sqrt{15})}} - \frac{1}{\cos{(\sqrt3)}} = \sqrt{1+15} - \sqrt{1+3} = 2$$ 
However, for the first part $\int^{\arctan{\sqrt{15}}}_{\arctan{\sqrt3}} \csc{(\theta)} d\theta$, I get 
$$ \int^{\arctan{\sqrt{15}}}_{\arctan{\sqrt3}} \csc{(\theta)} d\theta = \log \tan{\frac{\theta}{2}} |^{arctan{\sqrt{15}}}_{arctan{\sqrt3}}$$
How can I continue it?
 A: $$I=\int \dfrac{\sqrt{x^2+1}}x=\int\dfrac{\sec^3t}{\tan t}dt=\int\dfrac{\sin t}{\cos^2t\sin^2t}dt$$
Set $\cos t=y$
Alternatively $I=\displaystyle\int\dfrac{\sqrt{x^2+1}}{x^2} x \ dx$
Set $\sqrt{x^2+1}=u,x^2+1=u^2,x\ dx =u \ du$
A: Hints:
$$\dfrac{\sec^3\theta}{\tan \theta }=\dfrac{1}{\sin \theta \cos^2\theta }=\dfrac{1}{\sin \theta (1-\sin^2\theta)}=\dfrac1{\sin \theta}+\dfrac{\sin \theta}{1-\sin^2\theta}=\text {cosec}\,\theta+\dfrac{\sin \theta}{\cos^2\theta}. $$
$\displaystyle \int\csc x\,\mathrm dx=\ln|\csc(x)-\cot(x)|+C$ and  $\displaystyle \csc x=\sqrt{1-\frac1 {\tan^2x}}$.
Can you solve it now?
A: Think of the identities that you have available:
$$\begin{align}\cos^2\theta+\sin^2\theta&=1\\
1+\tan^2\theta&=\sec^2\theta\\
\cot^2\theta+1&=\csc^2\theta\\
\cosh^2\theta-\sinh^2\theta&=1\\
1-\tanh^2\theta&=\text{sech}^2\,\theta\\
\coth^2\theta-1&=\text{csch}^2\,\theta\end{align}$$
Of the $4$ that give you a formula for $1+x^2$, the one that seems to work best in the present context is the last. Accordingly we let $x=\text{csch}\,\theta$, $\sqrt{1+x^2}=\coth\theta$, $dx=-\text{csch}\,\theta\coth\theta\,d\theta$, so
$$\begin{align}\int\frac{\sqrt{1+x^2}}xdx&=-\int\coth^2\theta\,d\theta=-\int\left(1+\text{csch}^2\,\theta\right)d\theta=-\theta+\coth\theta+C\\
&=-\sinh^{-1}\left(\frac1x\right)+\sqrt{1+x^2}+C\\
&=-\ln\left(\frac1x+\sqrt{1+\frac1{x^2}}\right)+\sqrt{1+x^2}+C\\
&=\ln x-\ln\left(1+\sqrt{1+x^2}\right)+\sqrt{1+x^2}+C\end{align}$$
So
$$\int_{\sqrt3}^{\sqrt{15}}\frac{\sqrt{1+x^2}}xdx=\ln\sqrt{15}-\ln5+4-\ln{\sqrt3}+\ln3-2$$
As promised in the original question.
