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I am learning to calculate the arc length by reading a textbook, and there is a question

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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?

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    $\begingroup$ Please check your last step. Note that $\frac1 {\cos x}=\sec x=\sqrt{1+\tan^2x} $. $\endgroup$ – Thomas Shelby Feb 18 '19 at 7:02
  • $\begingroup$ @ThomasShelby Hi with your help, I solve the second part, but the first part I get stuck. Are there any equations so that I get replace $\tan{\frac{\theta}{2}}$ with $\tan{\theta}$ $\endgroup$ – GoingMyWay Feb 18 '19 at 7:49
  • $\begingroup$ @ThomasShelby I mean the result of $\tan{\frac{\arctan{\theta}}{2}}$ $\endgroup$ – GoingMyWay Feb 18 '19 at 7:58
  • $\begingroup$ $\tan{\frac{\theta}{2}}=\csc\theta-\cot\theta$. Now $\csc\theta=\sqrt{1-\frac1 {\tan^2\theta}} $. $\endgroup$ – Thomas Shelby Feb 18 '19 at 10:06
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    $\begingroup$ @ThomasShelby Thanks, $\csc{\theta}=\sqrt{\frac{\tan^2{\theta}+1}{\tan^2{\theta}}}$. Now I get the right answer, could you please add your hint in your answer, I will accept your answer later. $\endgroup$ – GoingMyWay Feb 18 '19 at 10:28
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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?

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  • $\begingroup$ Hi, I updated the question following your hint. $\endgroup$ – GoingMyWay Feb 18 '19 at 4:49
  • $\begingroup$ $\int\csc x dx$ $\endgroup$ – Thomas Shelby Feb 18 '19 at 6:49
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$$I=\int \dfrac{\sqrt{x^2+1}}x=\int\dfrac{\sec^3t}{\tan t}dt=\int\dfrac{\sin t}{\cos^2t\sin^2t}dt$$

Set $\cos t=y$

Alternatively $I=\displaystyle\int\dfrac{\sqrt{x^2+1}}{x^2} x \ dx$

Set $\sqrt{x^2+1}=u,x^2+1=u^2,x\ dx =u \ du$

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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.

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