Integration of $\frac{1}{x^2-a^2}$ by trigonometric substitution?

Question

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

Now, I know this can be done by splitting the function into two integrable functions,

$\displaystyle\dfrac{1}{2a}\int \bigg(\dfrac{1}{x-a} - \dfrac{1}{x+a}\bigg)dx$

And then doing the usual stuff.

My question is, how can we do this by using trigonometric substitution?

The only thing that gets in my mind is $x=a\sec\theta$, but then got stuck on proceeding further.

Any help would be appreciated.

Answer 1

We have $$I=\int \frac{1}{x^2-a^2}dx$$

Let $x=a\cos\theta$, then $dx=-a\sin\theta d\theta$.

\begin{align} I &= \int \frac{1}{x^2-a^2}dx \\ &= \int \frac{-a\sin\theta}{a^2(\cos^2\theta-1)}d\theta \\ &=\int \frac{-a\sin\theta}{-a^2\sin^2\theta}d\theta \\ &= \frac1a\int \csc\theta d\theta \\ &= -\frac1a\ln\left\lvert\csc\theta+\cot\theta\right\rvert +C\\ &= -\frac1a\ln\left\lvert\frac{1}{\sin\theta}+\frac{\cos\theta}{\sin\theta}\right\rvert +C\\ &= -\frac1a\ln\left\lvert\frac{1+x/a}{\pm\sqrt{1-x^2/a^2}}\right\rvert +C\\ &= -\frac1a\ln\left\lvert\frac{\sqrt{1+x/a}\sqrt{1+x/a}}{\sqrt{1-x/a}\sqrt{1+x/a}}\right\rvert +C\\ &= -\frac1{2a}\ln\left\lvert\frac{1+x/a}{1-x/a}\right\rvert +C\\ &= -\frac1{2a}\ln\left\lvert\frac{a+x}{a-x}\right\rvert +C\\ \end{align}

Answer 2

In addition to other answers: for $x > a$ a substitution $$x = a \cosh(x)$$ seems appropriate. The rest of the solution should work just the same.

Answer 3

You might use $x=a\sec\theta$: $dx=a\sec\theta\tan\theta\,d\theta$ and the integral becomes $$ \int\frac{1}{a^2(\sec^2\theta-1)}a\sec\theta\tan\theta\,d\theta = \frac{1}{a}\int\frac{\cos^2\theta}{\sin^2\theta}\frac{1}{\cos\theta}\frac{\sin\theta}{\cos\theta}\,d\theta = \frac{1}{a}\int\frac{1}{\sin\theta}\,d\theta $$ and this is a known fellow: set $\theta=2u$, so the integral becomes (leaving aside the factor $1/a$): $$ \int\frac{1}{\sin u\cos u}\,du= \int\frac{1}{\cos^2u+\sin^2u}{\sin u\cos u}\,du= \int\frac{\cos u}{\sin u}\,du+\int\frac{\sin u}{\cos u}\,du $$ You thus get $$ \log\lvert\sin u\rvert-\log\lvert\cos u\rvert+c= \log\left\lvert\tan\frac{\theta}{2}\right\rvert+c $$ Now back substitute.

Oh, well, but if we do $x=a\cos t$, instead? $$ \int\frac{1}{a^2(\cos^2t-1)}(-a\sin t)\,dt= \frac{1}{a}\int\frac{1}{\sin t}\,dt $$ Speedier, isn't it? Anyway, not easier than observing that $$ \frac{1}{x^2-a^2}=\frac{1}{2a}\left(\frac{1}{x-a}-\frac{1}{x+a}\right) $$