Question

How do you find the indefinite integral of $\int {\sec \dfrac{x}{2}dx}$?

Answer 1

In solving the given question, we have to integrate the given function, first multiply and divide with $\tan \dfrac{x}{2} + \sec \dfrac{x}{2}$, then consider the denominator as a variable i.e.,$u$, then differentiate both sides, and then further simplification we will get the required result.

Complete step by step solution:
Indefinite integral refers to an integral that does not have any upper and lower limit.
Given function is $\int {\sec \dfrac{x}{2}dx}$,
Now multiply and divide the given function with $\tan \dfrac{x}{2} + \sec \dfrac{x}{2}$, we get,
$\Rightarrow \int {\sec \dfrac{x}{2}\left( {\dfrac{{\tan \dfrac{x}{2} + \sec \dfrac{x}{2}}}{{\tan \dfrac{x}{2} + \sec \dfrac{x}{2}u}}} \right)dx}$,
Now simplifying by multiplying we get,
$\Rightarrow \int {\dfrac{{\sec \dfrac{x}{2}\tan \dfrac{x}{2} + {{\sec }^2}\dfrac{x}{2}}}{{\tan \dfrac{x}{2} + \sec \dfrac{x}{2}u}}dx}$,
Now let us consider $\tan \dfrac{x}{2} + \sec \dfrac{x}{2} = u$,
Now differentiate both sides we get,
$\Rightarrow \dfrac{d}{{dx}}u = \dfrac{d}{{dx}}\left( {\tan \dfrac{x}{2} + \sec \dfrac{x}{2}} \right)$,
Now distributing the differentiation, we get,
$\Rightarrow \dfrac{{du}}{{dx}} = \dfrac{d}{{dx}}\tan \dfrac{x}{2} + \dfrac{d}{{dx}}\sec \dfrac{x}{2}$,
Now applying derivative formulas we get,
$\Rightarrow \dfrac{{du}}{{dx}} = \dfrac{1}{2}{\sec ^2}\dfrac{x}{2} + \dfrac{1}{2}\sec \dfrac{x}{2}\tan \dfrac{x}{2}$,
Now take $dx$to the right hand side we get,
$\Rightarrow 2du = \left( {{{\sec }^2}\dfrac{x}{2} + \sec \dfrac{x}{2}\tan \dfrac{x}{2}} \right)dx$,
Now substituting the values we get,
$\Rightarrow \int {\dfrac{{2du}}{u}}$,
Now using the integration formula $\int {\dfrac{1}{x}dx} = \ln x + c$, we get,
$\Rightarrow \int {\dfrac{{2du}}{u}} = 2\ln \left| u \right|$,
We know that $u = \tan \dfrac{x}{2} + \sec \dfrac{x}{2}$, now substituting the value in the above we get,
$\Rightarrow \int {\sec \dfrac{x}{2}} = 2\ln \left| {\tan \dfrac{x}{2} + \sec \dfrac{x}{2}} \right| + C$.
So, the integral for the given function is $2\ln \left| {\tan \dfrac{x}{2} + \sec \dfrac{x}{2}} \right| + C$.

Final Answer:
$\therefore$The integral for the given function $\int {\sec \dfrac{x}{2}dx}$ will be equal to $2\ln \left| {\tan \dfrac{x}{2} + \sec \dfrac{x}{2}} \right| + C$.

Note:
An indefinite is an integral that does not contain the upper and lower limit. Indefinite integral is also known as anti-derivative or Prime integral. Indefinite integral of a function f is generally a differentiable function F whose derivative is equal to the original function f. It is represented as,
$\int {f\left( x \right)dx} = F\left( x \right) + C$,
Some of the important formulas that we use while solving integration problems are given below:
$\int {dx = x + c}$,
$\int {adx = ax + c}$,
$\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}}} + c$,
$\int {\dfrac{1}{x}dx} = \ln x + c$,
$\int {{e^x}dx = {e^x} + c}$,
$\int {{a^x}dx = \dfrac{{{a^x}}}{{\ln a}} + c}$,
$\int {\sin xdx = - \cos x + c}$,
$\int {\cos xdx = \sin x + c}$,
$\int {{{\sec }^2}xdx = \tan x + c}$,
$\int {{{\csc }^2}xdx = - \cot x + C}$,
$\int {\sec x\tan xdx = \sec x + C}$,
$\int {\csc x\cot xdx = - \csc x + C}$,
$\int {\dfrac{1}{{\sqrt {1 - {x^2}} }}dx = {{\sin }^{ - 1}}x + c}$,
$\int {\dfrac{1}{{1 + {x^2}}}dx = {{\tan }^{ - 1}}x + c}$.