Question

What is the antiderivative of $\sec x$?

Answer 1

We know that antiderivative means we have to do the integration of the given function. In the above problem, we are asked to find the antiderivative of $\sec x$ means we have to find the integration of $\sec x$ with respect to x. To find the integration of $\sec x$ with respect to x, we are going to take the derivative of $\sec x+\tan x$ then we are going to arrange this result of the derivative in such a manner so that we will get the integration of $\sec x$.

Complete step-by-step solution:
In the above problem, we are asked to find the antiderivative of $\sec x$ which means we have to find the integration of $\sec x$.
Let us find the derivative of $\sec x+\tan x$ with respect to x.
$\dfrac{d\left( \sec x+\tan x \right)}{dx}=\sec x\tan x+{{\sec }^{2}}x$
Now, taking $\left( \sec x+\tan x \right)$ as common in the R.H.S of the above equation we get,
$\Rightarrow \dfrac{d\left( \sec x+\tan x \right)}{dx}=\sec x\left( \tan x+\sec x \right)$
As you can see that $\left( \sec x+\tan x \right)$ is common in both the sides of the above equation so we can take $\left( \sec x+\tan x \right)$ as “u” in the above equation then the above equation will look like:
$\Rightarrow \dfrac{d\left( u \right)}{dx}=\sec x\left( u \right)$
Rearranging the above equation we get,
$\dfrac{du}{u}=\sec xdx$
Integrating on both the sides of the above equation we get,
$\int{\dfrac{du}{u}}=\int{\sec xdx}$
We know that integration of $\dfrac{1}{u}$ with respect to u is $\ln u$ then writing this form in the integration form we get,
$\ln u+C=\int{\sec xdx}$
Substituting the value of “u” as $\left( \sec x+\tan x \right)$ in the above equation we get,
$\ln \left( \sec x+\tan x \right)+C=\int{\sec xdx}$
Hence, we have found the antiderivative of $\sec x$ as $\ln \left( \sec x+\tan x \right)+C$.

Note: You can check the antiderivative of $\sec x$ which we have solved above is correct or not by doing the differentiation on this antiderivative. And if the result of the differentiation is the same as $\sec x$ then the antiderivative which we have solved above is correct.
The antiderivative of $\sec x$ is equal to $\ln \left( \sec x+\tan x \right)+C$. Now, taking the derivative of $\ln \left( \sec x+\tan x \right)+C$ with respect to x we get,
$\dfrac{d\ln \left( \sec x+\tan x \right)}{dx}=\dfrac{1}{\sec x+\tan x}\left( \sec x\tan x+{{\sec }^{2}}x \right)$
Taking $\sec x$ as common from the R.H.S of the above equation and we get,
$\dfrac{d\ln \left( \sec x+\tan x \right)}{dx}=\dfrac{\sec x}{\sec x+\tan x}\left( \tan x+\sec x \right)$
Now, $\left( \sec x+\tan x \right)$ will be cancelled out from the numerator and the denominator of the R.H.S of the above equation we get,
$\begin{aligned} & \dfrac{d\ln \left( \sec x+\tan x \right)}{dx}=\dfrac{\sec x}{1}\left( 1 \right) \\\ & \Rightarrow \dfrac{d\ln \left( \sec x+\tan x \right)}{dx}=\sec x \\\ \end{aligned}$
As you can see that the differentiation of the antiderivative of $\sec x$ is $\sec x$ so the antiderivative which we have solved above is correct.