7
$\begingroup$

So I was recently looking at a new book on Calculus[1] that I got and decided to look at the techniques for integration that it listed. I then decided after a while to make my own integral and evaluate it using the techniques that it listed. Here is the integral I came up with[2]:$$\int_1^{9}\dfrac{dx}{x\sqrt{81-x^2}}$$which I thought that I might be able to evaluate. Here is my attempt at doing so:


We use the substitution $x=9\sin(\theta)\implies\theta=\sin^{-1}\left(\dfrac x9\right)$, and then we can have $dx=9\cos(\theta)d\theta$, and then $\sqrt{81-x^2}$ becomes[3]$$\sqrt{81-81\sin^2(\theta)}=9\sqrt{\sin^2\theta}\implies9\sqrt{\cos^2(\theta)}\implies9|\cos(\theta)|=\sqrt{81-\dfrac{x^2}9}$$$$\because|\cos(\theta)|=\sqrt{a^2-\dfrac{x^2}a}\because\text{the inverse sine function}$$$$\text{oscillates between }-\dfrac\pi2\text{ and }\dfrac\pi2\text{ which implies}$$$$\cos(\theta)=|\cos(\theta)|=\sqrt{a^2-\dfrac{x^2}a}=\sqrt{81-\dfrac{x^2}9}$$This implies the integral $I(t)$ is now$$\int_1^9\dfrac{\require{cancel}\cancel{9\cos(\theta)}d\theta}{9\sin(\theta)\cancel{(9\cos(\theta))}}$$$$=\dfrac19\int_1^9\csc(\theta)d\theta$$Now there’s only one problem: Here’s what we get when we evaluate the integral:$$\dfrac19\left[\operatorname{Ln}\left(\sin\left(\dfrac\theta2\right)\right)-\operatorname{Ln}\left(\cos\left(\dfrac\theta2\right)\right)\right]_1^9$$There’s only one problem. I know how to evaluate any $\sin(\dfrac x2)$, but I don’t know how to evaluate any $\cos(\dfrac x2)$. Then, I decided to go back in the book to the part on trigonometric functions and saw definition ($16.13$):

($16.13$)$\quad\cos^2(u)=\dfrac{1+\cos u}2$

This was great because now I could finally integrate the function. Then,$$\dfrac19\left[\operatorname{Ln}\left(\sin\left(\dfrac\theta2\right)\right)-\operatorname{Ln}\left(\cos\left(\dfrac\theta2\right)\right)\right]_1^9$$$$=\dfrac19\left[\ln\left(\sqrt{\dfrac12(1-\cos(\theta))}\right)-\ln\left(\sqrt{\dfrac12(1+\cos(\theta))}\right)\right]_1^9$$$$\dfrac1{18}\left[\ln\left(\dfrac{1-\cos(\theta)}2\right)-\ln\left(\dfrac{1+\cos(\theta)}2\right)\right]_1^9$$$$\implies\dfrac1{18}\left[\ln\left(\dfrac{1-\cos(\theta)}{1+\cos(\theta)}\right)\right]_1^9$$$$\implies\dfrac1{18}\left[\ln\left(\dfrac{9-\sqrt{81-x^2}}{9+\sqrt{81-x^2}}\right)\right]_1^9$$which converges to$$-\dfrac1{18}\ln\left(\dfrac{(9-\sqrt{80})^2}{161}\right)$$or approximately $0.603108$


My question


Did I evaluate the integral correctly, or what could I do to evaluate it correctly?


Notes


[1]Calculus by Elliot Mendelson, PHD.

[2]Based off of Chapter $32$ “Techniques of Integration II” example $32.12$

[3]This is done using Strategy II: “If $\sqrt{a^2-x^2}$ occurs in an integrated, try the substitution $x=a\sin\theta$.”

$\endgroup$
4
  • 2
    $\begingroup$ The limits for $\theta$ are not $1$ and $9,$ but $\sin^{-1}(1/9)$ and $\pi/2.$ $\endgroup$ Aug 21, 2023 at 15:12
  • 2
    $\begingroup$ The final denominator should be $1$ instead of $161.$ $\endgroup$ Aug 21, 2023 at 15:20
  • 1
    $\begingroup$ Apart from this, everything is correct. wolframalpha.com/… The result is approximately $0.32081.$ google.com/search?q=-%28ln%289-sqrt%2880%29%29%29%2F9 $\endgroup$ Aug 21, 2023 at 15:27
  • 1
    $\begingroup$ fwiw, your computations include a term of the form $$\log\frac{1-z}{1+z}$$, which would suggest that the inverse hyperbolic tangent is a good candidate for finding the closed form antiderivative for your initial expression. for example, see the derivative taken in WolframAlpha here. $\endgroup$ Aug 21, 2023 at 15:41

2 Answers 2

4
$\begingroup$

Fundamentally most of this is correct.


The biggest error comes in the very last computation: $$ -\left. \dfrac1{18}\left[\ln\left(\dfrac{9-\sqrt{81-x^2}}{9+\sqrt{81-x^2}}\right)\right] \right|_{x=1} \ne -\dfrac1{18}\ln\left(\dfrac{(9-\sqrt{80})^2}{161}\right) $$ To see this, note that $$ \left. \dfrac{9-\sqrt{81-x^2}}{9+\sqrt{81-x^2}} \right|_{x=1} = \frac{9-\sqrt{80}}{9+\sqrt{80}} $$ which I'm sure you realized. Multiplying by $9-\sqrt{80}$ on top and bottom to rationalize, we get $$ \frac{9-\sqrt{80}}{9+\sqrt{80}} \cdot \frac{9-\sqrt{80}}{9-\sqrt{80}} = \frac{(9-\sqrt{80})^2}{9^2 - (\sqrt{80})^2} = \frac{(9-\sqrt{80})^2}{81-80} = (9-\sqrt{80})^2 $$ (whereas you added on bottom). One can then simplify to conclude a final answer of $$ - \frac 1 9 \ln \left(9 - \sqrt{80}\right) $$ Note that WolframAlpha agrees (up to rounding error and algebraic manipulation, since it chooses a slightly different form).


Throughout, you should change your bounds after shifting to $\theta$ as well. For instance, $$ \int_1^{9}\dfrac{dx}{x\sqrt{81-x^2}} \ne \dfrac19\int_1^9\csc(\theta)d\theta $$ but instead $$ \int_1^{9}\dfrac{dx}{x\sqrt{81-x^2}} \ne \dfrac19\int_{\arcsin(1/9)}^{\pi/2} \csc(\theta)d\theta $$ This also prevents the need to convert back to $x$. Of course, if one wishes to insist on this style of writing, they should at least specify the variable in the bounds, if just for clarity and bookkeeping purposes. For instance, $$ \int_1^{9}\dfrac{dx}{x\sqrt{81-x^2}} = \dfrac19\int_{x=1}^{x=9} \csc(\theta)d\theta $$ is an equality I would take less issue with.

One can say essentially the same about the case of evaluations.


Anecdotally, I would suggest getting very familiar with the \begin{align*} ... \end{align*} environment in MathJax/LaTeX, because it would make your post and your code a lot cleaner.

$\endgroup$
0
$\begingroup$

I agree with PrincessEev that

$$ I=\frac{1}{9} \int_{\sin ^{-1} \frac{1}{9}}^{\frac{\pi}{2}} \csc \theta d \theta $$

and try to give a complete solution to it using the result that $$ \int \csc \theta d \theta=-\ln |\csc \theta+\cot \theta|+C $$ then we have $$ \begin{aligned} I & =-\frac{1}{9}[\ln |\csc \theta+\cot \theta|]_{\sin ^{-1} \frac{1}{9}}^{\frac{\pi}{2}} \\ & =-\frac{1}{9}[\ln |1+0|-\ln |9+4 \sqrt{5}|] \\ & =\frac{1}{9} \ln (9+4 \sqrt{5}) \end{aligned} $$ Wish it helps!

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .