Question

Integrate ${{\sec }^{3}}x$ with respect to x.

Answer 1

Hint: We will be using the concept of integral calculus to solve the problem. We will be using the integration by parts method to integrate ${{\sec }^{3}}x$.

Complete step-by-step answer:
Now, we have been given $\int{{{\sec }^{3}}xdx}$ and we have to find its value.
Now, we will be using integration by parts to solve the problem. We know that,
$\int{f\left( x \right)g\left( x \right)dx}=f\left( x \right)\int{g\left( x \right)dx}-\int{\left( f'\left( x \right)\int{g\left( x \right)dx} \right)dx}$
So, we split $\int{{{\sec }^{3}}xdx}$ as,
$I=\int{{{\sec }^{2}}x}\sec xdx$
So, using integral by parts we have,
\[\begin{align}
  & I=\sec x\int{{{\sec }^{2}}xdx-\int{\sec x\tan x\int{{{\sec }^{2}}xdx}}} \\
 & I=\sec x\int{{{\sec }^{2}}xdx-\int{\left( \sec x\tan x\int{{{\sec }^{2}}xdx} \right)}}dx..........\left( 1 \right) \\
\end{align}\]
Now, we have to find $\int{{{\sec }^{3}}xdx}$.
Now, we know that $\dfrac{d}{dx}\left( \tan x \right)={{\sec }^{2}}x$ and we know that differentiation and integration are reverse processes. So, we can find $\int{{{\sec }^{2}}xdx}$ by finding the anti – derivative of ${{\sec }^{2}}x$ which is tan x. Therefore,
$\int{{{\sec }^{2}}x}dx=\tan x$
Now, we have from (1),
$I=\sec x\tan x-\int{\sec x{{\tan }^{2}}xdx}$
Now, we know that ${{\sec }^{2}}x-{{\tan }^{2}}x=1$
${{\tan }^{2}}x={{\sec }^{2}}x-1$
Therefore, we have,
$\begin{align}
  & I=\sec x\tan x-\int{\sec x\left( {{\sec }^{2}}x-1 \right)dx} \\
 & I=\sec x\tan x-\int{\left( {{\sec }^{3}}x-\sec x \right)dx} \\
 & I=\sec x\tan x-\int{{{\sec }^{3}}xdx+\int{\sec xdx}} \\
\end{align}$
Now, we have taken $I=\int{{{\sec }^{3}}xdx}$. So, substituting this we have,
$\begin{align}
  & I=\sec x\tan x-I+\int{\sec xdx} \\
 & 2I=\sec x\tan x+\int{\sec xdx} \\
\end{align}$
Now, we know that,
$\int{\sec xdx=\log \left| \sec x+\tan x \right|+c}$
Therefore, we have,
$\begin{align}
  & 2I=\sec x\tan x+\log \left| \sec x+\tan x \right|+c \\
 & I=\dfrac{1}{2}\left( \sec x\tan x \right)+\dfrac{1}{2}\log \left| \sec x+\tan x \right|+c \\
\end{align}$

Note: To solve these types of questions it is important to note that we have re-substituted $\int{{{\sec }^{3}}xdx=I}$ in our solution to further simplify it. Also it is advisable to remember standard integral like,
$\begin{align}
  & \int{{{\sec }^{2}}xdx=\tan x+c} \\
 & \int{\sec xdx=\log \left| \sec x+\tan x \right|+c} \\
\end{align}$