MHB Integral using trig substitution

[tmt1]

I have this integral:

$$\int_{}^{}\frac{1}{x^2 - 9} \,dx$$

I believe I can use trig substitution with this so I can set $x = 3 sec\theta$

Evaluating this, I get

$$ln|\csc\left({\theta}\right) - \cot\left({\theta}\right)| + C$$

Since $x^2 - 9 = 9sec^2\theta - 9$, then $\frac{x^2 - 9}{3} = tan^2\theta$ and $tan\theta = \sqrt{\frac{x^2 - 9}{3}}$.

Then we know $csc\theta = \frac{sec\theta}{tan\theta}$ and in this context $sec\theta = x / 3$, so $csc\theta = \frac{x / 3}{ \sqrt{\frac{x^2 - 9}{3}}}$.

Thus the answer is

$$ln| \frac{x / 3}{ \sqrt{\frac{x^2 - 9}{3}}} - \sqrt{\frac{3}{x^2 - 9}}| + C$$

Is this correct?

 

[Greg]

You can avoid a trig sub:

$$\int\dfrac{1}{x^2-9}\,dx=\dfrac16\int\left(\dfrac{1}{x-3}-\dfrac{1}{x+3}\right)\,dx=\dfrac16\left(\ln|x-3|-\ln|x+3|\right)+C$$

$$=\dfrac16\ln\left|\dfrac{x-3}{x+3}\right|+C$$

 

[soroban]

\int \frac{dx}{x^2-9}

Let x \,=\,3\sec\theta \quad\Rightarrow\quad dx \;=\; 3\sec\theta\tan\theta\,d\theta
x^2-9 \:=\:9\sec^2\theta-9 \;=\;9(\sec^2\theta-1) \;=\;9\tan^2\theta

\text{Substitute: }\;\int\frac{3\sec\theta\tan\theta\,d\theta}{9\tan^2\theta} \;=\;\tfrac{1}{3}\int\frac{\sec\theta}{\tan\theta}d\theta \;=\;\tfrac{1}{3}\int\frac{d\theta}{\sin\theta}

. . =\;\tfrac{1}{3}\int\csc\theta\,d\theta \;=\;\tfrac{1}{3}\ln|\csc\theta - \cot\theta| + CBack-substitute: .\sec\theta \:=\:\frac{x}{3} \quad\Rightarrow\quad \csc\theta \:=\:\frac{x}{\sqrt{x^2-9}} \quad\Rightarrow\quad \cot\theta \:=\:\frac{3}{\sqrt{x^2-9}}

We have: . \tfrac{1}{3}\ln \left|\frac{x}{\sqrt{x^2-9}} - \frac{3}{\sqrt{x^2-9}}\right| + C <br /> \;=\;\tfrac{1}{3}\ln\left|\frac{x-3}{\sqrt{x^2-9}}\right| +C

. . =\;\tfrac{1}{3}\ln\left|\frac{x-3}{\sqrt{(x-3)(x+3)}}\right| + C \;=\;\tfrac{1}{3}\ln\left|\frac{\sqrt{x-3}}{\sqrt{x+3}}\right| + C \;=\;\tfrac{1}{3}\ln\left|\sqrt{\frac{x-3}{x+3}}\right| +C

. . =\;\tfrac{1}{6}\ln\left|\frac{x-3}{x+3}\right| + C

 

[tmt1]

You can avoid a trig sub:

$$\int\dfrac{1}{x^2-9}\,dx=\dfrac16\int\left(\dfrac{1}{x-3}-\dfrac{1}{x+3}\right)\,dx=\dfrac16\left(\ln|x-3|-\ln|x+3|\right)+C$$

$$=\dfrac16\ln\left|\dfrac{x-3}{x+3}\right|+C$$

How does $\frac{1}{x^2-9}$ simplify to $\frac{1}{6}( \frac{1}{x-3}-\frac{1}{x+3})$? I see that $x^3 - 9 = (x - 3)(x + 3)$ but how is this derived?

 

[MarkFL]

How does $\frac{1}{x^2-9}$ simplify to $\frac{1}{6}( \frac{1}{x-3}-\frac{1}{x+3})$? I see that $x^3 - 9 = (x - 3)(x + 3)$ but how is this derived?

We use partial fraction decomposition. We begin with the factorization:

$$\frac{1}{x^2-9}=\frac{1}{(x+3)(x-3)}$$

Then the assumption this factored form can also be written in the following form (sum):

$$\frac{1}{(x+3)(x-3)}=\frac{A}{x+3}+\frac{B}{x-3}$$

Now, I would at this point, for simplicity since we have linear factors, use the Heaviside cover-up method. On the left side, we see the root of the first factor is:

$$x=-3$$

Then we cover-up that factor, and evaluate what's left using that value for $x$ and equate that to $A$:

$$A=\frac{1}{-3-3}=-\frac{1}{6}$$

Then we do the same thing with the other factor for $B$:

$$B=\frac{1}{3+3}=\frac{1}{6}$$

And so, we obtain:

$$\frac{1}{(x+3)(x-3)}=\frac{-\dfrac{1}{6}}{x+3}+\frac{\dfrac{1}{6}}{x-3}=\frac{1}{6}\left(\frac{1}{x-3}-\frac{1}{x+3}\right)$$

 

[soroban]

Back when I was a student, one of the standard forms we learned was:

. . \int\frac{du}{u^2-a^2} \;=\;\frac{1}{2a}\ln\left|\frac{u-a}{u+a}\right| + C

It occupied only a few brain cells and saved hours of time.