# Integration changing limits - does the question have an error?

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

• Fist things first, your substitution should be $x=\sec \theta$, since $x=\sec x$ is an equation. – Jaideep Khare Sep 15 '17 at 20:07
• @JaideepKhare typo, fixed! – vik1245 Sep 15 '17 at 20:08
• I have edited further. Looks good now. – Jaideep Khare Sep 15 '17 at 20:13
• the result should be $$2/3\,\sqrt {3}-3/4\,\sqrt {2}$$ – Dr. Sonnhard Graubner Sep 15 '17 at 20:16
• 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. – 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.

• Now finish the solution! – 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 $$\frac{2\sqrt{2}}{3}-\frac{\sqrt{3}}{2}$$