# How do you use substitution to integrate [ 1 / ( t^2 + t - 2)^(1/2)]dt?

Jun 20, 2015

You can see the final answer here.

This is going to end up using the following major pieces of knowledge:

• Completing the Square
• Trigonometric Substitution
• The trick to $\int \sec \theta d \theta$ (involves u-substitution)
• Properties of Logarithms

$\int \frac{1}{\sqrt{{t}^{2} + t - 2}} \mathrm{dt}$

${t}^{2} + t - 2 = \left(t - 1\right) \left(t + 2\right)$

Although you CAN factor this, I'm going to complete the square instead. I want to get rid of the square root.

${t}^{2} + t - 2 = {t}^{2} + t + \frac{1}{4} - \frac{1}{4} - 2 = {\left(t + \frac{1}{2}\right)}^{2} - \frac{9}{4}$

$\implies \int \frac{1}{\sqrt{{\left(t + \frac{1}{2}\right)}^{2} - \frac{9}{4}}} \mathrm{dt}$

Surprisingly, now we have the form:

$\sqrt{{x}^{2} - {a}^{2}}$

Now things are looking up for Trigonometric Substitution. I was taught to use $x = a \text{trig} \theta$, so with $x = t + \frac{1}{2}$ and $a = \frac{3}{2}$, let:
$t + \frac{1}{2} = \frac{3}{2} \sec \theta$
${\left(t + \frac{1}{2}\right)}^{2} = \frac{9}{4} {\sec}^{2} \theta$
$\mathrm{dt} = \frac{3}{2} \sec \theta \tan \theta d \theta$
$\sqrt{{\left(t + \frac{1}{2}\right)}^{2} - \frac{9}{4}} = \sqrt{\frac{9}{4} {\sec}^{2} \theta - \frac{9}{4}} = \frac{3}{2} \sqrt{{\sec}^{2} \theta - 1} = \frac{3}{2} \sqrt{{\tan}^{2} \theta} = \frac{3}{2} \tan \theta$

Now, we have:

$\implies \int \frac{1}{\cancel{\frac{3}{2}} \cancel{\tan \theta}} \cancel{\frac{3}{2}} \sec \theta \cancel{\tan \theta} d \theta$

$= \int \sec \theta d \theta$

Not so bad, I guess... but you have to know this integral by now. Not because I expect that you should know more, but you'll see it eventually if you haven't already (it's pretty popular...). At least, I've been taught to do this integral. If you haven't learned it yet, here is how to do $\int \sec \theta d \theta$:

With malice aforethought... multiply by $\frac{\sec \theta + \tan \theta}{\sec \theta + \tan \theta}$.

$= \int \frac{\sec \theta \left(\sec \theta + \tan \theta\right)}{\sec \theta + \tan \theta} d \theta$

$= \int \frac{{\sec}^{2} \theta + \sec \theta \tan \theta}{\sec \theta + \tan \theta} d \theta$

Now, let:
$u = \sec \theta + \tan \theta$
$\mathrm{du} = \sec \theta \tan \theta + {\sec}^{2} \theta d \theta$

$\implies \int \frac{1}{u} \mathrm{du}$

WOAH! Much easier now!

$= \ln | u |$

$= \ln | \sec \theta + \tan \theta | + C$

Going back to our substitutions:

$\sec \theta = \frac{2}{3} \left(t + \frac{1}{2}\right)$
$\tan \theta = \frac{2}{3} \sqrt{{\left(t + \frac{1}{2}\right)}^{2} - \frac{9}{4}} = \frac{2}{3} \sqrt{{t}^{2} + t - 2}$

So the final result is:

$= \ln | \frac{2}{3} \left(t + \frac{1}{2}\right) + \frac{2}{3} \sqrt{{t}^{2} + t - 2} | + C$

$= \ln | \frac{1}{3} \left(2 \left(t + \frac{1}{2}\right) + 2 \sqrt{{t}^{2} + t - 2}\right) | + C$

$= \ln | \frac{1}{3} \left(2 t + 1 + 2 \sqrt{{t}^{2} + t - 2}\right) | + C$

Now, with the properties of logarithms, separate $\ln$ out.

$= \ln | 2 \sqrt{{t}^{2} + t - 2} + 2 t + 1 | + \ln \left(\frac{1}{3}\right) + C$

but $\ln \left(\frac{1}{3}\right) = - \ln 3$, both of which are just constants. So now, we get a new constant $D$ which we can embed into $C$ and call it $C$ again... oh well. It's all still a constant, and the point of it is to say that there are infinite vertical shifts of the antiderivative that can be differentiated to give $\frac{1}{\sqrt{{t}^{2} + t - 2}}$. In other words, the value of the constant is not relevant. Finally, we get:

$= \textcolor{b l u e}{\ln | 2 \sqrt{{t}^{2} + t - 2} + 2 t + 1 | + C}$