Question

What is the integral of ${{\tan }^{2}}\left( x \right)\sec \left( x \right)$ ?

Answer 1

Hint: Here we will use some standard equations to solve. We know that the integral of $\sec x$ is $\ln \left( \left| \sec \left( x \right)+\tan \left( x \right) \right| \right)$ .
From trigonometry we have ${{\tan }^{2}}x={{\sec }^{2}}x-1$ .
The Integration by parts is given by $\int{uv=u\int{v-\int{du\int{v}}}}$ .
Using the above results, we will find the integral.

Complete step by step answer:
We have been given that
${{\tan }^{2}}\left( x \right)\sec \left( x \right)$
To write the integral of given function.
We will assume the value of function is equal to $I$ and then solve further:
$I=\int{\left( {{\tan }^{2}}\left( x \right)\sec \left( x \right) \right)}dx$ …(1)
From the trigonometric identity we have ${{\tan }^{2}}x={{\sec }^{2}}x-1$ thus substituting this value in equation (1), we get:
\[I=\int{\left( {{\sec }^{2}}x-1 \right)\sec xdx}\] …(2)
Now multiply $\sec x$ to both the terms of bracket from equation (2), we get:
$I=\int{\left( {{\sec }^{3}}x-\sec x \right)dx}$
$\Rightarrow \int{{{\sec }^{3}}xdx-\int{\sec xdx}}$ …(3)
Now we know that the integral of $\sec x$ is $\ln \left( \left| \sec \left( x \right)+\tan \left( x \right) \right| \right)$ therefore applying that to equation (3) we get:
$I=\int{{{\sec }^{3}}xdx-\ln \left( \left| \sec \left( x \right)+\tan \left( x \right) \right| \right)}$ …(4)
Now we will find the integral of ${{\sec }^{3}}x$ by applying integration of parts in which let us consider $u=\sec x,v={{\sec }^{2}}x$
Therefore, we can write integral of ${{\sec }^{3}}x$ as shown below:
$\int{{{\sec }^{3}}xdx=\int{\left( \sec x \right)}}\left( {{\sec }^{2}}x \right)dx$ …(5)
Now apply integration by parts $\int{uv=u\int{v-\int{du\int{v}}}}$ to equation (5) we get:
$\begin{align}
  & \int{{{\sec }^{3}}xdx=\sec x\tan x-\int{\tan x\left( \sec x\tan x \right)dx}} \\
 & \Rightarrow \sec x\tan x-\int{{{\tan }^{2}}x\left( \sec x \right)dx} \\
\end{align}$
Now once again we use the trigonometric identity ${{\tan }^{2}}x={{\sec }^{2}}x-1$ for the above equation, we get:
$\Rightarrow \sec x\tan x-\int{\sec x\left( {{\sec }^{2}}x-1 \right)dx}$ ….(6)
Now multiply the terms in equation (6) we get:
$\Rightarrow \sec x\tan x-\left( \int{{{\sec }^{3}}xdx-\int{\sec xdx}} \right)$ …(7)
Now we know that the integral of $\sec x$ is $\ln \left( \left| \sec \left( x \right)+\tan \left( x \right) \right| \right)$ therefore using this in equation (7) we get:
$\Rightarrow \sec x\tan x-\int{{{\sec }^{3}}xdx}+\ln \left( \left| \sec \left( x \right)+\tan \left( x \right) \right| \right)$ …(8)
Now we can transfer the term $\int{{{\sec }^{3}}xdx}$ to the right-hand side in equation (8) we get:
$\begin{align}
  & \int{{{\sec }^{3}}xdx}+\int{{{\sec }^{3}}xdx}=\sec x\tan x+\ln \left( \left| \sec \left( x \right)+\tan \left( x \right) \right| \right) \\
 & \Rightarrow \dfrac{1}{2}\left( \sec x\tan x+\ln \left( \left| \sec \left( x \right)+\tan \left( x \right) \right| \right) \right) \\
\end{align}$
Now substituting this value in equation (4) we have:
$I=\dfrac{1}{2}\left( \sec x\tan x+\ln \left( \left| \sec \left( x \right)+\tan \left( x \right) \right| \right) \right)-\ln \left( \left| \sec \left( x \right)+\tan \left( x \right) \right| \right)$
Thus, on solving we have the integral of ${{\tan }^{2}}\left( x \right)\sec \left( x \right)$ is given by:
\[\int{{{\tan }^{2}}x\sec xdx=I=\dfrac{1}{2}\left( \sec x\tan x-\ln \left( \left| \sec \left( x \right)+\tan \left( x \right) \right| \right) \right)+c}\]

Note: We can find the integral of $\sec x$ by using the integration by parts which can be as shown below:
$\int{\sec xdx}$
We multiply and divide by the term $\dfrac{\sec x+\tan x}{\sec x+\tan x}$ and then we select therefore we can write:
$\begin{align}
  & \int{\sec x\dfrac{\sec x+\tan x}{\sec x+\tan x}dx} \\
 & \Rightarrow \int{\dfrac{{{\sec }^{2}}x+\sec x\tan x}{\sec x+\tan x}dx} \\
\end{align}$
$u=\sec x+\tan x,du=\sec x\tan x+{{\sec }^{2}}xdu,v=1$ we get the new expression as :
$\int{\dfrac{du}{u}}$
We know the integral of $\int{\dfrac{du}{u}}$ is $\ln \left| u \right|+c$
Therefore, we have:
$\int{\sec xdx}=\ln \left| \sec x+\tan x \right|+c$