I am asking about changing the limits of integration.

I have the following integral to evaluate -

$$\int_2^{3}\frac{1}{(x^2-1)^{\frac{3}{2}}}dx$$ using the substitution $x = sec \theta$.

The problem states

Use the substitution to change the limits into the form $\int_a^b$ where $a$ and $b$ are multiples of $\pi$.

Now, this is what I did.

$$ x= \sec \theta$$ $$\frac{dx}{d\theta} = \sec\theta \tan\theta$$ $$dx = sec\theta tan\theta \ d\theta$$

$$\begin{align}\int_2^{3}\frac{1}{(x^2-1)^{\frac{3}{2}}}\,dx \\ &= \int\frac{1}{(\sec^2\theta-1)^{\frac{3}{2}}}\sec\theta \tan\theta \,d\theta \\ &= \int\frac{\sec\theta \tan\theta}{(\tan^2\theta)^{\frac{3}{2}}} \,d\theta \\ &= \int\frac{\sec\theta \tan\theta}{\tan^3\theta} \, d\theta \\ &= \int\frac{\sec\theta}{\tan^2\theta} \, d\theta \\ &= \int\frac{\cos\theta}{\sin^2\theta} \, d\theta \\ &= \int \csc\theta \cot\theta \, d\theta \end{align}$$

But here is my problem. I know that when $x = 2$, $$2 = \sec \theta$$ $$\frac{1}{2} = \cos \theta$$ $$\frac{\pi}{3} = \theta$$

but when $x = 3$ $$3 = \sec \theta$$ $$\frac{1}{3} = \cos \theta$$ $$\arccos\left(\frac{1}{3}\right) = \theta = $$ but this does not give me a definite result in $\pi$. The book says the following -

enter image description here

where $\arccos\left(\frac{1}{3}\right) = \frac{\pi}{3}$ Am I the only one or is the book wrong in this instance?

  • $\begingroup$ Fist things first, your substitution should be $x=\sec \theta$, since $x=\sec x$ is an equation. $\endgroup$ – Jaideep Khare Sep 15 '17 at 20:07
  • $\begingroup$ @JaideepKhare typo, fixed! $\endgroup$ – vik1245 Sep 15 '17 at 20:08
  • $\begingroup$ I have edited further. Looks good now. $\endgroup$ – Jaideep Khare Sep 15 '17 at 20:13
  • 1
    $\begingroup$ the result should be $$2/3\,\sqrt {3}-3/4\,\sqrt {2}$$ $\endgroup$ – Dr. Sonnhard Graubner Sep 15 '17 at 20:16
  • 1
    $\begingroup$ If the book were right, the integral would be equal to 0. (as both limits are equal) You are right in that $\arccos\left(\frac 1 3\right)$ is not a multiple of pi, so I would suggest an error in the textbook. $\endgroup$ – George Coote Sep 15 '17 at 20:21

The book is indeed incorrect.

$$arccos(\frac{1}{3}) \neq \frac{\pi}{3}$$

Note if it was the case that $arccos(\frac{1}{3}) = \frac{\pi}{3}$ as stated in the solutions,

then the integral would total 0.

  • $\begingroup$ Now finish the solution! $\endgroup$ – Nosrati Sep 15 '17 at 20:24

For simplicity, write $a=\arccos\frac{1}{2}$ and $b=\arccos\frac{1}{3}$. One can take care of them at the end.

It is true that $a=\pi/3$, but it's definitely wrong that $b=\pi/3$. Actually, $b\approx1.230959$, but you won't need that.

The integral becomes $$ \int_a^b-\frac{1}{(\sec^2\theta-1)^{3/2}}\sec^2\theta\sin\theta\,d\theta $$ Now it's better to write the integrand in terms of sine and cosine, recalling that the interval of integration is contained in $(0,\pi/2)$. Thus $$ \sec^2\theta-1= \frac{1-\cos^2\theta}{\cos^2\theta}= \frac{\sin^2\theta}{\cos^2\theta} $$ so the integrand is $$ -\frac{\cos^3\theta}{\sin^3\theta}\frac{1}{\cos^2\theta}\sin\theta= -\frac{\cos\theta}{\sin^2\theta} $$ Thus we get $$ \int_2^{3}\frac{1}{(x^2-1)^{3/2}}\,dx= \int_a^b-\frac{\cos\theta}{\sin^2\theta}\,d\theta= \int_a^b-\frac{1}{\sin^2\theta}\,d(\sin\theta)= \left[\frac{1}{\sin\theta}\right]_a^b $$ Since $a=\arccos(1/2)$ and $b=\arccos(1/3)$, we get $$ \sin a=\sqrt{1-\cos^2a}=\frac{\sqrt{3}}{2} \qquad \sin b=\sqrt{1-\cos^2b}=\frac{2\sqrt{2}}{3} $$ with no “sign uncertainty”, because $a$ and $b$ lie in $(0,\pi/2)$.

Thus the integral is $$ \frac{2\sqrt{2}}{3}-\frac{\sqrt{3}}{2} $$


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.