Find the integral \int \frac{\sec^2 x}{25 - \tan^2 x} \,dx

Question:

Find the integral

{eq}\int \frac{\sec^2 x}{25 - \tan^2 x} \,dx {/eq}

Indefinite Integration:


The integrals that are evaluated regardless of the limits are known as indefinite Integrals. After integrating the indefinite integrals, we add an arbitrary constant known as the Constant of Integration.

The formula used to evaluate this integral is {eq}\displaystyle\int \dfrac{1}{x}dx=\log|x|+C {/eq}.

Answer and Explanation:


Given: {eq}I=\displaystyle\int \dfrac{\sec^2 x}{25 - \tan^2 x} \,dx {/eq}

Let {eq}\tan x =u\Rightarrow \sec^2x \ dx =du {/eq}

{eq}\Rightarrow I=-\dfrac{du}{u^2-25}\\\Rightarrow I=-\dfrac{du}{(u+5)(u-5)}\\\Rightarrow I=\dfrac{1}{10}\left(\displaystyle\int\dfrac{1}{u-5}du-\displaystyle \int\dfrac{1}{u+5}du\right)\\\Rightarrow I=\dfrac{1}{10}(\log|u-5|-\log|u+5|)+C {/eq}

Now, substituting back {eq}u=\tan x {/eq}, we get:-

{eq}\Rightarrow I=\dfrac{1}{10}(\log|\tan x-5|-\log|\tan+5|)+C {/eq}


Learn more about this topic:

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Indefinite Integrals as Anti Derivatives

from Math 104: Calculus

Chapter 12 / Lesson 11
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