# How do you integrate int 1/sqrt(3x-12sqrtx-21)  using trigonometric substitution?

May 24, 2018

$\frac{2}{\sqrt{3}} \left(x - 4 \sqrt{x} - 7\right) + \frac{4}{\sqrt{3}} \ln \left(\sqrt{x} + 2 + \sqrt{x - 4 \sqrt{x} - 7}\right) + C$

#### Explanation:

$\int \frac{\mathrm{dx}}{\sqrt{3 x - 12 \sqrt{x} - 21}}$

I've added the differential $\mathrm{dx}$ to your integrand. While it may seem like a fanciful calculus accessory, it's actually very important, especially when doing trig sub problems.

First, let's get this into a more "normal" form for trig subs. Let $x = {t}^{2}$. Note this implies that $\mathrm{dx} = 2 t \mathrm{dt}$ and that $\sqrt{x} = t$.

$= \int \frac{2 t \mathrm{dt}}{\sqrt{3 {t}^{2} - 12 t - 21}}$

Now complete the square in the denominator:

$= 2 \int \frac{t}{\sqrt{3 \left({t}^{2} - 4 t + 4\right) - 21 - 12}} \mathrm{dt}$

$= 2 \int \frac{t}{\sqrt{3 {\left(t - 2\right)}^{2} - 33}} \mathrm{dt}$

$= \frac{2}{\sqrt{3}} \int \frac{t}{\sqrt{{\left(t - 2\right)}^{2} - 11}} \mathrm{dt}$

This is optional, but now I'd let $s = t - 2$. This implies that $\mathrm{ds} = \mathrm{dt}$ and that $t = s + 2$.

$= \frac{2}{\sqrt{3}} \int \frac{s + 2}{\sqrt{{s}^{2} - 11}} \mathrm{ds}$

Now, to clear the denominator, I'd let $s = \sqrt{11} \sec \theta$.

My reason for this substitution is that ${s}^{2} - 11 = 11 {\sec}^{2} \theta - 11 = 11 \left({\sec}^{2} \theta - 1\right) = 11 {\tan}^{2} \theta$. This will clear out our denominator.

Moreover, note that $\mathrm{ds} = \sqrt{11} \sec \theta \tan \theta d \theta$. Proceeding:

$= \frac{2}{\sqrt{3}} \int \frac{\sqrt{11} \sec \theta + 2}{\sqrt{11 {\tan}^{2} \theta}} \left(\sqrt{11} \sec \theta \tan \theta d \theta\right)$

$= \frac{2}{\sqrt{3}} \int \frac{\sqrt{11} \sec \theta + 2}{\sqrt{11} \tan \theta} \left(\sqrt{11} \sec \theta \tan \theta d \theta\right)$

$= \frac{2}{\sqrt{3}} \int \left(\sqrt{11} \sec \theta + 2\right) \sec \theta d \theta$

$= \frac{2}{\sqrt{3}} \int \left(\sqrt{11} {\sec}^{2} \theta + 2 \sec \theta\right) d \theta$

These are two fairly well known integrals. You may need to look the second one up.

$= \frac{2}{\sqrt{3}} \left(\sqrt{11} \tan \theta + 2 \ln \left(\left\mid \sec \theta + \tan \theta \right\mid\right)\right)$

Recall that $s = \sqrt{11} \sec \theta$, so $\sec \theta = \frac{s}{\sqrt{11}}$. Moreover, $\tan \theta = \sqrt{{\sec}^{2} \theta - 1} = \sqrt{{s}^{2} / 11 - 1} = \sqrt{\frac{{s}^{2} - 11}{11}}$. Then:

$= 2 \sqrt{\frac{11}{3}} \sqrt{\frac{{s}^{2} - 11}{11}} + \frac{4}{\sqrt{3}} \ln \left(\left\mid \frac{s}{\sqrt{11}} + \sqrt{\frac{{s}^{2} - 11}{11}} \right\mid\right)$

Do some more simplifications. Remember the log rule $\log \left(A / B\right) = \log \left(A\right) - \log \left(B\right)$.

$= \frac{2}{\sqrt{3}} \sqrt{{s}^{2} - 11} + \frac{4}{\sqrt{3}} \ln \left(\left\mid s + \sqrt{{s}^{2} - 11} \right\mid\right) - \frac{4}{\sqrt{3}} \ln \sqrt{11}$

Now use $s = t - 2$ and $t = \sqrt{x}$, so $s = \sqrt{x} - 2$;

$= \frac{2}{\sqrt{3}} \left({\left(\sqrt{x} - 2\right)}^{2} - 11\right) + \frac{4}{\sqrt{3}} \ln \left(\left\mid \sqrt{x} - 2 + \sqrt{{\left(\sqrt{x} - 2\right)}^{2} - 11} \right\mid\right) - \frac{4}{\sqrt{3}} \ln \sqrt{11}$

$= \frac{2}{\sqrt{3}} \left(x - 4 \sqrt{x} - 7\right) + \frac{4}{\sqrt{3}} \ln \left(\sqrt{x} + 2 + \sqrt{x - 4 \sqrt{x} - 7}\right) + C$

I've added the constant of integration, into which the $\frac{4}{\sqrt{3}} \ln \sqrt{11}$ term has been absorbed.

The absolute value bars are also unnecessary because the argument of the natural log function is, by necessity, greater than $2$ and always positive.