Question

How do you integrate $\int{{{\sec }^{2}}x\tan x}$ ?

Answer 1

Hint: Here in this problem, we need to perform integration of the given trigonometric expression. There are various integration formulae which we will be using here. Apart from that, Pythagoras identities of trigonometry can also be used. We will solve using the substitution method in integration.

Complete step-by-step solution:
Let’s begin to solve the problem.
Some important integration rules are:
$\begin{align}
  & \Rightarrow \int{1dx=x+C} \\
 & \Rightarrow \int{adx=ax+C} \\
 & \Rightarrow \int{{{x}^{n}}dx=\dfrac{{{x}^{n+1}}}{n+1}+C} \\
 & \Rightarrow \int{\sin xdx=-\cos x+C} \\
 & \Rightarrow \int{\cos xdx=\sin x+C} \\
 & \Rightarrow \int{{{\sec }^{2}}xdx=\tan x+C} \\
 & \Rightarrow \int{\cos e{{c}^{2}}xdx=-\cot x+C} \\
 & \Rightarrow \int{\sec x\left( \tan x \right)dx=\sec x+C} \\
 & \Rightarrow \int{\cos ecx\left( \cot x \right)dx=-\cos ecx+C} \\
\end{align}$
Some differentiation rules are also used.
$\begin{align}
  & \Rightarrow \dfrac{d}{dx}\sec x=\sec x\tan x \\
 & \Rightarrow \dfrac{d}{dx}\tan x={{\sec }^{2}}x \\
\end{align}$
Now, write the expression which needs to be integrated.
$I=\int{{{\sec }^{2}}x\tan x dx}......(i)$
Let u = tanx
Differentiate on both sides by using $\dfrac{d}{dx}\tan x={{\sec }^{2}}x$, we will get:
$\Rightarrow du={{\sec }^{2}}xdx$
Replace all the values in equation(i) with new values by substituting ‘du’ and ‘u’ in equation(i).
$I=\int{udu}$
Integrate with respect to du by using $\int{{{x}^{n}}dx=\dfrac{{{x}^{n+1}}}{n+1}+C}$, we get:
$\Rightarrow \dfrac{{{u}^{2}}}{2}+C$
Again substitute the value of ‘u’ in above equation we get:
$\Rightarrow \dfrac{{{\tan }^{2}}x}{2}+C$
This is the final answer.

Note: There is an alternative method to solve this question. That method will be similar to the above method. Let’s discuss it also.
$I=\int{{{\sec }^{2}}x\tan x}$
First break ${{\sec }^{2}}x$ into $\sec x$ and $\sec x$ like this:
$I=\int{\sec x\tan x\sec xdx}......(i)$
Now, let u = secx
Differentiate both sides using $\dfrac{d}{dx}\sec x=\sec x\tan x$, we will get:
$\Rightarrow du=\sec x\tan xdx$
Now, substitute ‘du’ and ‘u’ in equation(i) we will get:
$I=\int{udu}$
Integrate with respect to du by using $\int{{{x}^{n}}dx=\dfrac{{{x}^{n+1}}}{n+1}+C}$, we get:
$\Rightarrow \dfrac{{{u}^{2}}}{2}+C$
Again substitute the value of ‘u’ in above equation we get:
$\Rightarrow \dfrac{{{\sec }^{2}}x}{2}+C$
This is another answer for the same question. For this question, the substitution method is the best approach and is mostly used in the questions of integration.