Question

How do you integrate $\int {\left[ {{{\left( {\sec \left( x \right)} \right)}^5}} \right]} dx$?

Answer 1

Hint: The above question is based on the concept of integration. Since it is an indefinite integral which has no upper and lower limits, we can apply integration properties by integrating it where the power increases by one and we can find the antiderivative of the above expression.

Complete step by step solution:
Integration is a way of finding the antiderivative of any function. It is the inverse of differentiation. It denotes the summation of discrete data. Calculation of small problems is an easy task but for adding big problems which include higher limits, integration methods are used. The above given expression is an indefinite integral which means there are no upper or lower limits given.
The above equation should be in the below form.
\[\int {f\left( x \right) = F\left( x \right) + C} \]
where C is constant.
So, the above expression is given:
$\int {\left[ {{{\left( {\sec \left( x \right)} \right)}^5}} \right]} dx$
The secant function can be split into different powers so that we can further integrate it by parts. So the power 5 can be split into 2 and 3.
\[{\sec ^5}x = {\sec ^2}x{\sec ^3}x\]
Now by integrating it we get,
\[\int {{{\sec }^5}x} = \int {{{\sec }^2}x{{\sec }^3}xdx} \]
Since we know that
\[\dfrac{d}{{dx}}\left( {\tan x} \right) = {\sec ^2}x\].
Therefore, by substituting the value we get,
\[
\int {{{\sec }^5}x} dx = \int {d(\tan x){{\sec }^3}x} \\
\int {{{\sec }^5}xdx = \tan x{{\sec }^3}x - \int {\tan xd\left( {{{\sec }^3}\left( x \right)} \right)} }
\\
\]
where
\[\dfrac{d}{{dx}}\left( {{{\sec }^3}\left( x \right)} \right) = 3{\sec ^2}\left( x \right)\dfrac{d}{{dx}}\sec
\left( x \right) = 3{\sec ^3}x\tan x\]
\[
\int {{{\sec }^5}x} = \tan x{\sec ^3}x - 3\int {\left( {{{\sec }^2}x - 1} \right){{\sec }^3}} \\

\int {{{\sec }^5}x = \tan x{{\sec }^3}x + 3\int {{{\sec }^3}xdx - 3\int {{{\sec }^5}xdx} } } \\
\]
The integral now appears on both sides of the equation and then solving it we obtain a reduction formula:
\[\int {{{\sec }^5}xdx = \dfrac{1}{4}\left( {\tan x{{\sec }^3}x + 3\int {{{\sec }^3}xdx} } \right)} \]
Now integrating secant function
\[
\int {{{\sec }^3}xdx = \int {\sec xd\left( {\tan x} \right)} } \\
\int {{{\sec }^3}xdx = \tan x\sec x - \int {{{\tan }^2}x\sec xdx} } \\
\int {{{\sec }^3}xdx = \tan x\sec x - \int {\left( {{{\sec }^2}x - 1} \right)\sec xdx} } \\
\int {{{\sec }^3}xdx = \dfrac{1}{2}\left( {\tan x\sec x + \int {\sec xdx} } \right)} \\
\]
By solving resulting integral we get,
\[\dfrac{d}{{dx}}\left( {\tan x + \sec x} \right) = \sec x(\tan x + \sec x)\]
So divide and multiply by \[\sec x + \tan x\]
\[
\int {\sec xdx = \int {\dfrac{{\sec x(\sec x + \tan x)}}{{\sec x + \tan x}} = \int {\dfrac{{\sec x(\sec x +
\tan x)}}{{\sec x + \tan x}}} } } \\
\int {\sec xdx = \ln |\sec x + \tan x| + C} \\
\]
Substituting all together
\[\int {{{\sec }^5}xdx = \dfrac{{2\tan x{{\sec }^3}x + 3\tan x\sec x + 3\ln |\sec x + \tan x|}}{8} + C} \]

Note: An important thing to note is that the trigonometric identity is used \[{\tan ^2}x = {\sec ^2}x - 1\]. The identity is formed in such a given way \[{\tan ^2}\theta = \dfrac{{{{\sin }^2}\theta }}{{{{\cos }^2}\theta }} = \dfrac{{1 - {{\cos }^2}\theta }}{{{{\cos }^2}\theta }} = {\sec ^2}\theta - 1\]. Therefore we substitute this identity for tangent function.