$$\int_{2}^{4} \frac{\sqrt{16-x^2}}{x}$$



$$\int_{2}^{4} \frac{\sqrt{16-x^2}}{x}=\int_{2}^{4} \frac{\sqrt{16-16\sin^2u}}{4\sin u}\cos u\,du=\int_{2}^{4} \frac{4\sqrt{1-\sin^2u}}{4\sin u}\cos u \,du=\int_{2}^{4} \frac{\cos^2u}{\sin u}du=\int_{2}^{4} \frac{1-\sin^2u}{\sin u}du=\int_{2}^{4} \frac{1}{\sin u}du-\int_{2}^{4} {\sin u}du$$

$$=\ln\left(\tan\left(\frac{u}{2}\right)\right)+\cos u$$

$\frac{x}{4}=\sin u$

$\tan u=\frac{\sqrt{x}}{x^2-16}$

$\ln\left(\frac{{x}}{8(\sqrt{x^2-16})}\right)+\left(\frac{\sqrt{16-x^2}}{4}\right)$ form $2$ to $4$

but I get a $\frac{4}{0}$ the end result according to Wolfram is $1.80$

  • $\begingroup$ I think u forgot to change the limits in this ntegration. When you take a substitution you need to change the limits. $\endgroup$ – Murtuza Vadharia Feb 18 '16 at 18:10
  • $\begingroup$ There's still one more mistake. When you took a substitution x=4sinu then dx=4cosudu $\endgroup$ – Murtuza Vadharia Feb 18 '16 at 18:12

If we write $$\frac{\sqrt{16-x^2}}{x} = \frac{x \sqrt{16-x^2}}{x^2},$$ then the substitution $u^2 = 16-x^2$, $x^2 = 16-u^2$, $x \, dx = -u \, du$ immediately yields $$\int_{x=2}^4 \frac{\sqrt{16-x^2}}{x} \, dx = \int_{u=\sqrt{12}}^0 \frac{-u^2}{16-u^2} \, du = \int_{u=0}^{\sqrt{12}} \frac{16}{16-u^2} - 1 \, dy.$$ Then we proceed by partial fraction decomposition, giving $$2 \int_{u=0}^{\sqrt{12}} \left(\frac{1}{4-u} + \frac{1}{4+u}\right) \, du - \sqrt{12} = 2 \log \frac{4+\sqrt{12}}{4-\sqrt{12}} - \sqrt{12},$$ which simplifies to $$4 \log (2 + \sqrt{3}) - 2 \sqrt{3}.$$

Since people seem to think that the above calculation is incorrect, here it is via trigonometric substitution along the same lines as attempted by the original question:

$$\begin{align*} \int_{x=2}^4 \frac{\sqrt{16-x^2}}{x} \, dx &= \int_{u = \pi/6}^{\pi/2} \frac{4 \sqrt{1 - \sin^2 u}}{4 \sin u} \cdot 4 \cos u \, du \\ &= 4 \int_{u=\pi/6}^{\pi/2} \frac{\cos^2 u}{\sin u} \, du \\ &= 4 \int_{u=\pi/6}^{\pi/2} \csc u - \sin u \, du \\ &= 4 \int_{u=\pi/6}^{\pi/2} \frac{\csc u (\csc u + \cot u)}{\csc u + \cot u} \, du - \left[ - 4\cos u \right]_{u=\pi/6}^{\pi/2} \\ &= 4\left[-\log|\csc u + \cot u| \right]_{u=\pi/6}^{\pi/2} - 2 \sqrt{3} \\ &= 4\log (2 + \sqrt{3}) - 2 \sqrt{3}.\end{align*}$$

And here is the WolframAlpha link.

  • $\begingroup$ Answers do not agree from above posts. Someone is wrong somewhere... $\endgroup$ – imranfat Feb 18 '16 at 18:28
  • $\begingroup$ @imranfat My answer is correct. Verify it yourself; I have done it using trigonometric substitution as well as in Mathematica. If I've been downvoted for a correct calculation, then I am more than happy to delete and leave everyone else ignorant as to the correct solution. $\endgroup$ – heropup Feb 18 '16 at 18:43
  • $\begingroup$ Just so you know I liked your answer so I up voted. I think you shouldn't delete. $\endgroup$ – randomgirl Feb 18 '16 at 18:47
  • $\begingroup$ I did not downvote. I just typed both answers in my TI, then I had to go. If Mathematica gave yours, then it must be ok... $\endgroup$ – imranfat Feb 18 '16 at 19:44


Multiply Numerator Denomerator by $x$ then substitute $u^2=16-x^2$


I think u forgot to change the limits and also factor of 4 is missing while putting dx term. Edited answer : Sorry the previous answer I made a slight mistake in taking limits enter image description here So our final answer will look like as seen in the photo above. As you can see that we get $tan(\pi/12)$ in the answer. By placing value of $tan(\pi/12)$ in different form. Look of the answer can be changed.

enter image description here

So now this is the correct answer. Apologies for slight mistake in limits in the previous answer.

  • $\begingroup$ the end result according to Wolfram is $1.80$ $\endgroup$ – gbox Feb 18 '16 at 18:29
  • 1
    $\begingroup$ I think you are also missing a factor of 4 towards the end of your work. (last 4 lines) $\endgroup$ – randomgirl Feb 18 '16 at 18:33
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    $\begingroup$ I think one of your limits is also off. $\endgroup$ – randomgirl Feb 18 '16 at 18:40
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    $\begingroup$ As i have shown in multiple ways, this accepted answer is incorrect. $\endgroup$ – heropup Feb 18 '16 at 18:56
  • 1
    $\begingroup$ @gbox You accepted a wrong answer... $\endgroup$ – imranfat Feb 18 '16 at 19:47

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