Evaluate {eq}\displaystyle \int ( \tan x + \cot x )^2 \ dx {/eq}

Question:

Evaluate {eq}\displaystyle \int ( \tan x + \cot x )^2 \ dx {/eq}

Integration of {eq}{\tan ^2}x {/eq}:

The integration of the trigonometric function {eq}{\tan ^2}x {/eq} is found out by simplifying the given function {eq}{\tan ^2}x {/eq} as {eq}{\tan ^2}x = {\sec ^2}x - 1 {/eq}. Now, we can directly find out the integration of {eq}f\left( x \right) = {\sec ^2}x - 1 {/eq} as follows {eq}\int {f\left( x \right)} = \int {\left( {{{\sec }^2}x - 1} \right)dx} = {\tan}x - x + C {/eq} where {eq}C {/eq} is the integrating constant.

Answer and Explanation: 1


The given integral is,

{eq}\int {{{\left( {\tan x + \cot x} \right)}^2}} {/eq}


Simplifying the trigonometric function,

{eq}\begin{align*} {\left( {\tan x + \cot x} \right)^2} &= {\tan ^2}x + {\cot ^2}x + 2\tan x\cot x\\ &= {\tan ^2}x + {\cot ^2}x + 2\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left( {\tan x\cot x = 1} \right)\\ &= \left( {{{\sec }^2}x - 1} \right) + \left( {{{\csc }^2}x - 1} \right) + 2\\ &= {\sec ^2}x + {\csc ^2}x \end{align*} {/eq}


The given integral can also be written as,

{eq}\int {{{\left( {\tan x + \cot x} \right)}^2}} = \int {\left( {{{\sec }^2}x + {{\csc }^2}x\;} \right)dx} {/eq}


Now, integrating the above integral with respect to x as,

{eq}\begin{align*} \int {{{\left( {\tan x + \cot x} \right)}^2}} dx &= \int {\left( {{{\sec }^2}x + {{\csc }^2}x\;} \right)dx} \\ &= \tan x - \cot x + C \end{align*} {/eq}


Thus, {eq}\int {{{\left( {\tan x + \cot x} \right)}^2}} dx = \tan x - \cot x + C {/eq}


Learn more about this topic:

Loading...
Integration Problems in Calculus: Solutions & Examples

from

Chapter 13 / Lesson 13
26K

Learn what integration problems are. Discover how to find integration sums and how to solve integral calculus problems using calculus example problems.


Related to this Question

Explore our homework questions and answers library