Trig Substitution Integral Question

Question

My class is going over trig substitution, but I can't figure this one out, mostly because it's not in the correct form. Could someone help explain how to set up this problem?

$$ \int \frac {1}{x^2-25} dx $$

Answer 1

You don't need trig here: use partial fractions.

Answer 2

If you insist on using trigonometric substitution, the correct one would be $x=5\sec\theta$. \begin{align*} \int \frac{dx}{x^2-25} &= \int\frac{\color{red}{5}\sec\theta\color{red}{\tan\theta}\,d\theta}{\color{red}{25\tan^2\theta}} = \frac{1}{5}\int\frac{\sec\theta}{\tan\theta}\,d\theta \\ &= \frac{1}{5}\int \csc\theta\,d\theta \\ &= -\frac{1}{5}\ln\left|\csc\theta + \cot\theta\right| + C \\ \end{align*} Drawing a right triangle with base angle $\theta$, hypotenuse $x$ and adjacent side $5$ (so that $\sec\theta = \frac{x}{5})$, we see that the opposite side must have length $\sqrt{x^2-25}$, so $\csc\theta = \frac{x}{\sqrt{x^2-25}}$ and $\cot\theta = \frac{5}{\sqrt{x^2-25}}$
Therefore \begin{align*} -\frac{1}{5}\ln\left|\csc\theta + \cot\theta\right| + C &= -\frac{1}{5}\ln\left|\frac{x}{\sqrt{x^2-25}} + \frac{5}{\sqrt{x^2-25}}\right| + C \\ &= -\frac{1}{5}\ln\left|\frac{x+5}{\sqrt{x^2-25}}\right| + C \\ &= -\frac{1}{5}\left(\ln\left|{x+5}\right|-\ln\left|{\sqrt{x^2-25}}\right|\right) + C \\ \\ &= -\frac{1}{5}\left(\ln\left|x+5\right|-\ln\left|\sqrt{x-5}\sqrt{x+5}\right|\right) + C \\ \\ &= -\frac{1}{5}\left(\ln\left|x+5\right|-\ln\left|\sqrt{x-5}\right|-\ln\left|\sqrt{x+5}\right|\right) + C \\ \\ &= -\frac{1}{5}\left(\ln\left|x+5\right|-\frac{1}{2}\ln\left|{x-5}\right|-\frac{1}{2}\ln\left|{x+5}\right|\right) + C \\ \\ &= -\frac{1}{5}\left(\frac{1}{2}\ln\left|x+5\right|-\frac{1}{2}\ln\left|{x-5}\right|\right) + C \\ \\ &= \frac{1}{10}\left(\ln|x-5| - \ln|x+5|\right) +C \end{align*}

This is definitely not the most straightforward solution but it's nice to know there's so inconsistency with the choice.

Answer 3

The integral $$ \int \frac {1}{x^2-25} dx $$ isn't in the correct form yet, but you can manipulate it easily $$ \int \frac {1}{x^2-25} dx = \int \frac {1}{25((\frac{x}{5})^2-1)} dx = \frac{1}{25}\int \frac {1}{(\frac{x}{5})^2-1} dx$$ now, just substitute $u=\frac{x}{5}, du=\frac{dx}{5}$ $$ \frac{1}{5}\int \frac {1}{u^2-1} du$$ continuing to manipulate... $$ \frac{1}{5}\int \frac {1}{u^2-1} du = \frac{1}{5}\int \frac {1}{\sec^2(\sec^{-1}u)-1} du=\frac{1}{5}\int \frac {1}{\tan^2(\sec^{-1}u)} du$$ From here on, it's just an inverse trigonometric substitution (albeit a messy one because it involves $\mid u \mid$). It would be far better to use inverse hyperbolic trigonometric substitution in this case, or as Robert Israel has already pointed out, partial fractions.