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

enter image description here

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?

  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ @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


$$\dfrac{\sec^3\theta}{\tan \theta }=\dfrac{1}{\sin \theta \cos^2\theta }=\dfrac{1}{\sin \theta (1-\sin^2\theta)}=\dfrac1{\sin \theta}+\dfrac{\sin \theta}{1-\sin^2\theta}=\text {cosec}\,\theta+\dfrac{\sin \theta}{\cos^2\theta}. $$

$\displaystyle \int\csc x\,\mathrm dx=\ln|\csc(x)-\cot(x)|+C$ and $\displaystyle \csc x=\sqrt{1-\frac1 {\tan^2x}}$.

Can you solve it now?

  • $\begingroup$ 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

$$I=\int \dfrac{\sqrt{x^2+1}}x=\int\dfrac{\sec^3t}{\tan t}dt=\int\dfrac{\sin t}{\cos^2t\sin^2t}dt$$

Set $\cos t=y$

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

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


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.