# Find the \int of \frac {1}{(4x^2-25)} using trig substitution. Give seen the \frac {1}{2} \ ln \...

## Question:

Find the {eq}\int of \frac {1}{(4x^2-25)} {/eq} using trig substitution. Give seen the {eq}\frac {1}{2} \ ln \ (2x+\sqrt{(4x^+25)}) {/eq} answer plenty of times but i need the trig-substitution version

## Integrals:

In order to evaluate the given integral, we will simplify the given function by applying the substitution method. In this particular case, we will make a trigonometric substitution, as it is mentioned above.

## Answer and Explanation:

Given {eq}\int \frac {1}{(4x^2-25)} \ dx {/eq}

Taking 4 common from denominator, we get:

{eq}\begin{align*} \ & = \frac{1}{4}\int \frac {1}{x^2-\frac{25}{4}} \ dx \end{align*} {/eq}

Put:

{eq}\begin{align*} \ & x = \frac{5}{2} \sec \theta \end{align*} {/eq}

Differentiating the above, we get:

{eq}\begin{align*} \ & dx = \frac{5}{2} \sec \theta \tan \theta \ d \theta \end{align*} {/eq}

Substituting the above values, we get:

{eq}\begin{align*} \ & = \frac{1}{4} \int \frac {1}{(\frac{25}{4} \sec^2 \theta-\frac{25}{4})} \cdot \frac{5}{2} \sec \theta \tan \theta \ d \theta \\ \\ \ & = \frac{1}{4} \cdot \frac{4}{25} \cdot \frac{5}{2}\int \frac {1}{ (\sec^2 \theta- 1 )} \cdot \sec \theta \tan \theta \ d \theta \\ \\ \ & = \frac{1}{10}\int \frac {1}{ \tan^2 \theta} \cdot \sec \theta \tan \theta \ d \theta \ \ \ \ \ \ \ \ \ \ \left[ (1+ \tan^2 x)= \sec^2 x \right] \\ \\ \ & = \frac{1}{10} \int \frac { \sec \theta}{ \tan \theta } \ d \theta \\ \\ \ & = \frac{1}{10} \int \frac { \frac{1}{\cos \theta}}{ \frac{\sin \theta}{\cos \theta} } \ d \theta \\ \\ \ & = \frac{1}{10} \int \frac{1}{\sin \theta} \ d \theta \\ \\ \ & = \frac{1}{10} \int \csc \theta \ d \theta \\ \\ \ & = \frac{1}{10} \ln |\csc \theta - \cot \theta| + c \end{align*} {/eq}

Substituting the value of {eq}\theta {/eq} back, we get:

{eq}\begin{align*} \ & = \frac{1}{10} \ln |\csc \left(\sec^{-1} \frac{2}{5}x \right) - \cot \left(\sec^{-1} \frac{2}{5}x \right)| + c \end{align*} {/eq}