4
$\begingroup$

My Work

$x = \tan\theta$

$dx = \sec^2\theta d\theta$

$\int \frac{\tan\theta\ln(\tan\theta)}{\sqrt{\tan^2\theta - 1}}\sec^2\theta d\theta$

$\int \sec\theta \tan\theta ln(tan\theta)$

$u = \ln(\tan\theta)$

$du = \frac{\sec^2 \theta}{\tan \theta}$

$dv = \sec \theta \tan \theta d\theta$

$v = \sec \theta$

$\sec\theta\ln\tan\theta - \int \frac{\sec^3\theta}{\tan \theta}$

$\sec\theta\ln\tan\theta - \int \sec^2\theta \csc\theta d\theta$

I'm stuck after here. Parts doesn't look particularly appealing. I don't see an easy substitution. Brain is pretty tired at this point. Anyone know what to do?

$\endgroup$
4
$\begingroup$

A start: Integrate by parts, letting $u=\ln x$ and $dv=\frac{x}{\sqrt{x^2-1}}\,dx$.

Remark: The substitution $x=\tan\theta$ is not good for $\sqrt{x^2-1}$. There the appropriate thing is $x=\sec\theta$ or $x=\cosh t$.

$\endgroup$
4
$\begingroup$

First integrate by parts, then substitute $x=\sec{\theta}$ $\implies \mathrm{dx}=\tan{\theta}\sec{\theta}\,\mathrm{d}\theta$:

$$\begin{align} \int\frac{x\,\ln{\left(x\right)}}{\sqrt{x^2-1}}\,\mathrm{d}x &=\sqrt{x^2-1}\,\ln{\left(x\right)}-\int\frac{\sqrt{x^2-1}}{x}\,\mathrm{d}x\\ &=\sqrt{x^2-1}\,\ln{\left(x\right)}-\int\frac{\sqrt{\sec^2{\theta}-1}}{\sec{\theta}}\,\tan{\theta}\sec{\theta}\,\mathrm{d}\theta\\ &=\sqrt{x^2-1}\,\ln{\left(x\right)}-\int\tan^2{\theta}\,\mathrm{d}\theta\\ &=\sqrt{x^2-1}\,\ln{\left(x\right)}-\int\left(\sec^2{\theta}-1\right)\,\mathrm{d}\theta\\ &=\sqrt{x^2-1}\,\ln{\left(x\right)}-\tan{\theta}+\theta+\color{grey}{constant}\\ &=\sqrt{x^2-1}\,\ln{\left(x\right)}-\sqrt{x^2-1}+\sec^{-1}{x}+\color{grey}{constant}\\ \end{align}$$

$\endgroup$
1
$\begingroup$

You can evaluate the integral without using secant substitution. Here is the way.

Since we know that \begin{gather*} \mathrm{d}\sqrt{x^2-1}=\frac{x}{\sqrt{x^2-1}}\mathrm{d} x, \end{gather*} we can calculate, by differentiation by parts, as follows, \begin{align*} &\quad \frac{x\ln(x)}{\sqrt{x^2-1}} \mathrm{d} x=\ln(x)\mathrm{d} \sqrt{x^2-1}=\mathrm{d} \big(\ln(x)\sqrt{x^2-1}\big)-\frac{\sqrt{x^2-1}}{x}\mathrm{d} x\\ &= \mathrm{d} \big(\ln(x)\sqrt{x^2-1}\big)-\frac{x^2-1}{x\sqrt{x^2-1}}\mathrm{d} x\\ &=\mathrm{d} \big(\ln(x)\sqrt{x^2-1}\big)-\frac{x}{\sqrt{x^2-1}}\mathrm{d} x+\frac{1}{x\sqrt{x^2-1}}\mathrm{d} x\\ &=\mathrm{d} \big(\ln(x)\sqrt{x^2-1}\big)-\mathrm{d}\big(\sqrt{x^2-1}\big)+\frac{1}{x^2\sqrt{1-\left(\frac{1}{x}\right)^2}}\mathrm{d} x\\ &=\mathrm{d} \big(\ln(x)\sqrt{x^2-1}\big)-\mathrm{d}\big(\sqrt{x^2-1}\big)-\frac{1}{\sqrt{1-\left(\frac{1}{x}\right)^ 2}}\mathrm{d} \left(\frac{1}{x}\right)\\ &=\mathrm{d}\left(\ln(x)\sqrt{x^2-1}-\sqrt{x^2-1}-\arcsin\left(\frac{1}{x}\right)\right), \end{align*} hence we have \begin{gather*} \int\frac{x\ln(x)}{\sqrt{x^2-1}}\mathrm{d} x=\ln(x)\sqrt{x^2-1}-\sqrt{x^2-1}-\arcsin\left(\frac{1}{x}\right)+C. \end{gather*}

$\endgroup$

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.