Question

Integrate dx/(25x²-4)½

Answer 1

The integral is given by:

$\int \frac{dx}{\sqrt{25x^2 - 4}} \\$

Let us assume that:

$u = \frac{5x}{2}$ (Equation 1)

$\Rightarrow du = \frac{5}{2} dx \\ \Rightarrow dx = \frac{2}{5} du \\$

From the (Equation 1), we can also write:

$x = \frac{2u}{5}$

Now:

Substituting $x = \frac{2u}{5}$ into the integration and integrating with respect to $du$ , we have:

$= \frac{2}{5} \int \frac{du}{\sqrt{4u^2 - 4}} \\ = \frac{2}{5} \int \frac{du}{2 \sqrt{u^2 - 1}} \\ = \frac{1}{5} \int \frac{du}{\sqrt{u^2 - 1}} \\$

Again:

Let us assume that:

$u = \sec (\theta ) \\ \Rightarrow du = \sec (\theta) \tan (\theta) \, d \theta \\$

Again:

Substituting $u = \sec (\theta)$ into the integration and integrating with respect to $d \theta$ , we have:

$= \frac{1}{5} \int \frac{\sec (\theta) \tan (\theta )}{\sqrt{\sec^2 (\theta) - 1}} \, d \theta \\$

$= \frac{1}{5} \int \frac{\sec (\theta) \tan (\theta)}{\tan (\theta)} \, d \theta$ [ $\because \sec^2 (\theta) - \tan^2 (\theta) = 1$ ]

$= \frac{1}{5} \int sec (\theta) \, d \theta \\ = \frac{1}{5} \ln \mid \sec (\theta) + \tan (\theta) \mid + C$

[ $\because \int \sec (x) \, dx = \ln \mid \tan (x) + \sec (x) \mid + C$ ( Where C is a constant of integration) ]

Undo substitution $\sec (\theta ) = u$ and $\tan (\theta) = \sqrt{u^2 -1}$ into the integration and we have:

$= \frac{1}{5} \ln \mid u + \sqrt{u^2 - 1} \mid + C$

Again:

Undo substitution $u = \frac{5x}{2}$ into the integration and we have:

$= \frac{1}{5} \ln \mid \frac{5x}{2} + \sqrt{ \frac{25x^2}{4} - 1 } \mid + C \\ = \frac{1}{5} \ln \mid \frac{5x}{2} + \frac{\sqrt{25x^2 -4}}{2} \mid + C \\$

$= \frac{1}{5} \ln \mid \frac{5x + \sqrt{25x^2 -4}}{2} \mid + C \\$

$= \frac{1}{5} \ln \mid 5x + \sqrt{25x^2 - 4} \mid - \frac{1}{5} \ln \mid 2 \mid + C$ $\left[ \because \ln \mid \frac{a}{b} \mid = \ln \mid a \mid - \ln \mid b \mid \right]$

$= \frac{1}{5} \ln \mid 5x + \sqrt{25x^2 - 4} \mid + K$ [ $\because K = C - \frac{1}{5} \ln \mid 2 \mid$ ( Where K is an arbitrary constant) ]