# How do you integrate int 1/(xsqrt(16x^2-9)) by trigonometric substitution?

Mar 18, 2018

$\int \frac{\mathrm{dx}}{x \sqrt{16 {x}^{2} - 9}} = \frac{x}{3 \left\mid x \right\mid} a r c \sec \left(\frac{4}{3} x\right) + C$

#### Explanation:

$\int \frac{\mathrm{dx}}{x \sqrt{16 {x}^{2} - 9}} = \int \frac{\mathrm{dx}}{3 x \sqrt{{\left(\frac{4}{3} x\right)}^{2} - 1}}$

Substitute:

$\frac{4}{3} x = \sec t$, $\mathrm{dx} = \frac{3}{4} \sec t \tan t \mathrm{dt}$

and restrict for now to $x \in \left(\frac{3}{4} , + \infty\right)$ so that $t \in \left(0 , \frac{\pi}{2}\right)$

$\int \frac{\mathrm{dx}}{x \sqrt{16 {x}^{2} - 9}} = \frac{3}{4} \int \frac{\sec t \tan t \mathrm{dt}}{\frac{9}{4} \sec t \sqrt{{\sec}^{2} t - 1}}$

$\int \frac{\mathrm{dx}}{x \sqrt{16 {x}^{2} - 9}} = \frac{1}{3} \int \frac{\tan t \mathrm{dt}}{\sqrt{{\sec}^{2} t - 1}}$

Use now the trigonometric identity:

${\sec}^{2} t - 1 = \frac{1}{\cos} ^ 2 t - 1 = \frac{1 - {\sin}^{2} t}{\cos} ^ 2 t = {\tan}^{2} t$

As $t \in \left(0 , \frac{\pi}{2}\right) \implies \tan t \ge 0$

then:

$\sqrt{{\sec}^{2} t - 1} = \tan t$

and:

$\int \frac{\mathrm{dx}}{x \sqrt{16 {x}^{2} - 9}} = \frac{1}{3} \int \frac{\tan t \mathrm{dt}}{\tan} t = \frac{1}{3} \int \mathrm{dt} = \frac{t}{3} + C$

and undoing the substitution:

$\int \frac{\mathrm{dx}}{x \sqrt{16 {x}^{2} - 9}} = \frac{1}{3} a r c \sec \left(\frac{4}{3} x\right) + C$

For $x \in \left(- \infty , - \frac{3}{4}\right)$ we can use the same procedure except that:

$\sqrt{{\sec}^{2} t - 1} = - \tan t$

and accordingly:

$\int \frac{\mathrm{dx}}{x \sqrt{16 {x}^{2} - 9}} = - \frac{1}{3} a r c \sec \left(\frac{4}{3} x\right) + C$

So we can express the integral for any $x$ as:

$\int \frac{\mathrm{dx}}{x \sqrt{16 {x}^{2} - 9}} = \frac{x}{3 \left\mid x \right\mid} a r c \sec \left(\frac{4}{3} x\right) + C$

Note that as:

$a r c \sec x = \arctan \left(x \sqrt{\frac{{x}^{2} - 1}{x} ^ 2}\right)$

We can also express the integral as:

$\int \frac{\mathrm{dx}}{x \sqrt{16 {x}^{2} - 9}} = \frac{x}{3 \left\mid x \right\mid} \arctan \left(\frac{4 x}{3} \sqrt{\frac{16 {x}^{2} - 9}{16 {x}^{2}}}\right) + C$

$\int \frac{\mathrm{dx}}{x \sqrt{16 {x}^{2} - 9}} = \frac{x}{3 \left\mid x \right\mid} \arctan \left(\frac{\sqrt{16 {x}^{2} - 9}}{3}\right) + C$