Question

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

Answer 1

Hint: 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:
\[\int {dx = x + c} \],
\[\int {adx = ax + c} \],
\[\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}}} + c\],
\[\int {\dfrac{1}{x}dx} = \ln x + c\],
\[\int {{e^x}dx = {e^x} + c} \],
\[\int {{a^x}dx = \dfrac{{{a^x}}}{{\ln a}} + c} \],
\[\int {\sin xdx = - \cos x + c} \],
\[\int {\cos xdx = \sin x + c} \],
\[\int {{{\sec }^2}xdx = \tan x + c} \],
\[\int {{{\csc }^2}xdx = - \cot x + C} \],
\[\int {\sec x\tan xdx = \sec x + C} \],
\[\int {\csc x\cot xdx = - \csc x + C} \],
\[\int {\dfrac{1}{{\sqrt {1 - {x^2}} }}dx = {{\sin }^{ - 1}}x + c} \],
\[\int {\dfrac{1}{{1 + {x^2}}}dx = {{\tan }^{ - 1}}x + c} \].