Question

How do you integrate $\int{{{\sec }^{2}}\left( \dfrac{x}{2} \right)\tan \left( \dfrac{x}{2} \right)dx}$ ?

Answer 1

Hint: Here we have to integrate the integral given. So we will use a substitution method to solve this integral. Firstly we will let $\tan \left( \dfrac{x}{2} \right)=u$ and differentiate it with respect to $x$ then we will substitute the values in the original integral and by using the basic variable integration formula we will simplify our integral. Finally we will put the left value in it and get our desired answer.

Complete answer:
We have to integrate the integral given as follows:
 $\int{{{\sec }^{2}}\left( \dfrac{x}{2} \right)\tan \left( \dfrac{x}{2} \right)dx}$….$\left( 1 \right)$
We will use the substitution method to solve the above integral.
Now Let us take,
$\tan \left( \dfrac{x}{2} \right)=u$…$\left( 2 \right)$
Differentiate both sides with respect to $x$ as follows:
$\Rightarrow \dfrac{d}{dx}\left( \tan \left( \dfrac{x}{2} \right) \right)=\dfrac{du}{dx}$…$\left( 3 \right)$
We know that the differentiation of tangent function is secant function square given as:$\dfrac{d}{dx}\tan x={{\sec }^{2}}x$
Using the above formula in equation (3) we get,
$\Rightarrow {{\sec }^{2}}\left( \dfrac{x}{2} \right)\times \dfrac{1}{2}=\dfrac{du}{dx}$
$\Rightarrow {{\sec }^{2}}\left( \dfrac{x}{2} \right)dx=2du$….$\left( 4 \right)$
On substituting the value from equation (2) and (4) in equation (1) we get,
$\Rightarrow \int{2udu}$
Now as we know that $\int{{{x}^{n}}dx=\dfrac{{{x}^{n+1}}}{n+1}}+C$ where $C$ is any constant using it above where $x=u$ and $n=1$ we get,
$\Rightarrow 2\times \dfrac{{{u}^{2}}}{2}+C$
$\Rightarrow {{u}^{2}}+C$
Replace the value from equation (2) above we get,
$\Rightarrow {{\tan }^{2}}\left( \dfrac{x}{2} \right)+C$
Where $C$ is any constant.
Hence the answer is $\int{{{\sec }^{2}}\left( \dfrac{x}{2} \right)\tan \left( \dfrac{x}{2} \right)dx}={{\tan }^{2}}\left( \dfrac{x}{2} \right)+C$ where $C$ is any constant.

Note:
Integration is also known as ant derivative or primitive function. When more than one trigonometric function is present inside the integral using a substitution method is the first way to solve the problem. In this question we can let the unknown variable equal to the secant value also and then solve the question accordingly but that will have more calculation. When there is no limit in the integral it is known as definite integral and we have to add a constant term in the final answer as that solution is true for any constant added to the value because the derivative of constant is $0$ .