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Q: Find the arc length of the curve $y=\ln(x)$ where $x$ ranges from $\sqrt{3}$ to $\sqrt{15}$.

I think I am stuck in calculation part.

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The answer is $2 + \ln(3) - \frac{1}{2}\ln(5)$. But I can't derive that from my last line.

help me, please.

  • 1
    Review your last two computations. $\sqrt{15}$ and $\sqrt{3}$ are values of $x$ and not for $\theta$ – 2017-01-19
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    You have to change the limits of the integration as well. – 2017-01-19
  • 0
    Does theta range from arctan(sqrt(3)) to arctan(sqrt(15))? The computations are still difficult to me.. – 2017-01-19
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    In your last term of the answer, is it really $-\frac{1}{2}\ln 5$? – 2017-01-19
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    Yes. My book says it is 2 + ln(3) - (1/2)*ln(5) units. – 2017-01-19
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    Just for suggestion. I will suggest you to start with$$F(x)=\int \frac{\sqrt{x^2+1}}{x}dx$$ – 2017-01-19
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    All your solutions after the substitution $x=\tan\theta$ are all wrong. Why? – 2017-01-19
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    When you make the substitution $x = \tan \theta$ you must recalculate your limits of integration to go with your new variable. (or convert back to x before applying the limits of integration. – 2017-01-19
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    When x = sqrt(3), theta = arctan(sqrt(3)). And when x = sqrt(15), theta = arctan(sqrt(15)). Are these right? – 2017-01-19
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    $\arctan \sqrt 3 = \frac \pi 3$ not that that is a big deal. What is more important to know is that $\csc \arctan \sqrt 3 = \frac 2{\sqrt 3}, \sec \arctan \sqrt 3 = 2,$ etc. – 2017-01-19
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    csc (arctan(sqrt(3)) = csc (pi/3) = 1/sin(pi/3) = 1/(sqrt(3)/2) = 2/sqrt(3). sec (arctan(sqrt(3)) = sec (pi/3) = 1/cos(pi/3) = 1/(1/2) = 2. What about sqrt(15)? Give me some hints, please. – 2017-01-19
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    @Kim It seems that your computations are going complicated. An alternate way of doing it is given in my answer below. Try it. We used the Fundamental Theorem of Calculus. – 2017-01-19
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    @Kim I will be going out for now and chat me if you have some more questions to raise. God bless. – 2017-01-19

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Your problem now is how to evaluate the integral $$\int_{\sqrt{3}}^{\sqrt{15}}\frac{\sqrt{x^2+1}}{x}dx.$$

Let $$F(x)=\int\frac{\sqrt{x^2+1}}{x}dx.$$ Let $x=\tan\theta$. All your computations after this substitution are all correct. Note that $$\csc\theta=\frac{\sqrt{x^2+1}}{x}$$ $$\cot\theta=\frac{1}{x}$$ and $$\sec\theta=\sqrt{x^2+1}.$$ Hence, we get $$F(x)=-\ln\left(\frac{\sqrt{x^2+1}+1}{x} \right)+\sqrt{x^2+1}+C.$$ Thus, $$ \begin{align} \int_{\sqrt{3}}^{\sqrt{15}}\frac{\sqrt{x^2+1}}{x}dx&=F(\sqrt{15})-F(\sqrt{3})\\ &=\left[-\ln\left(\frac{5}{\sqrt{15}}\right)+4+C\right]-\left[-\ln\left(\frac{3}{\sqrt{3}}\right)+2+C\right]\\ &=\ln\left(\frac{3}{\sqrt{3}}\right)-\ln\left(\frac{5}{\sqrt{15}}\right) +2\\ &=\ln\left[\frac{3}{\sqrt{3}}\div\frac{5}{\sqrt{15}}\right]+2\\ &=\ln\left[\frac{3}{\sqrt{3}} \cdot \frac{\sqrt{15}}{5}\right]+2\\ &=\ln\left[\frac{3}{1} \cdot \frac{\sqrt{5}}{5}\right]+2\\ &=\ln\left[\frac{3}{1} \cdot \frac{1}{\sqrt{5}}\right]+2\\ &=\ln\left(\frac{3}{\sqrt{5}}\right)+2\\ &=\ln 3-\ln\sqrt{5}+2\\ &=\ln 3-\frac{1}{2}(\ln 5)+2. \end{align}$$