# Find the following integral. {eq}\displaystyle \int \frac{\sqrt{(\tan x)^2 - 2}}{\cos^2 x} \ dx {/eq}

## Question:

Find the following integral.

{eq}\displaystyle \int \frac{\sqrt{(\tan x)^2 - 2}}{\cos^2 x} \ dx {/eq}

## Integration by Substitution:

In calculus, integration or anti-differentiation works opposite operation what differentiation does. In the substitution method of integration, the derivative of the substituted functions is already given in the integrand. It is used to rewrite the complex integrand into a simpler expression that can be easily integrated.

Some integral formulas for special functions are given below:

{eq}\begin{align*} \int {\sqrt {{x^2} - {a^2}} dx} &= \dfrac{x}{2}\sqrt {{x^2} - {a^2}} - \dfrac{{{a^2}}}{2}\ln \left| {x + \sqrt {{x^2} - {a^2}} } \right| + C\\ \int {\sqrt {{x^2} + {a^2}} dx} &= \dfrac{x}{2}\sqrt {{x^2} + {a^2}} + \dfrac{{{a^2}}}{2}\ln \left| {x + \sqrt {{x^2} + {a^2}} } \right| + C\\ \int {\sqrt {{a^2} - {x^2}} dx} &= \dfrac{x}{2}\sqrt {{a^2} - {x^2}} + \dfrac{{{a^2}}}{2}{\sin ^{ - 1}}\left( {\dfrac{x}{a}} \right) + C \end{align*} {/eq}

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Given Data:

• The given integral is: {eq}I = \int {\dfrac{{\sqrt {{{\left( {\tan x} \right)}^2} - 2} }}{{{{\cos }^2}x}}dx} {/eq}.

Rewrite the...