Question

What is the antiderivative of ${{\sec }^{2}}(x)$?

Answer 1

From the question given we have to find the antiderivative of ${{\sec }^{2}}(x)$. Generally, antiderivatives are opposite to the derivatives (inverse derivatives). We know that the derivative of$~\tan (x)$ is ${{\sec }^{2}}(x)$
We need to find the antiderivative of ${{\sec }^{2}}(x)$. Antiderivative means integral. From this we will get the antiderivative of ${{\sec }^{2}}(x)$.

Complete step by step solution:
Generally, antiderivatives are opposite to the derivatives (inverse derivatives).
We know that the derivative of$~\tan (x)$ is ${{\sec }^{2}}(x)$
We need to find the antiderivative of ${{\sec }^{2}}(x)$.
\Rightarrow $$$$\dfrac{d}{dx}\left( \tan x \right)={{\sec }^{2}}(x)
From the above equation it is clear that the derivative of $~\tan (x)$ is ${{\sec }^{2}}(x)$.
We know that the antiderivatives are inverse derivatives of the derivatives.
So, it is very clear that the antiderivative of the ${{\sec }^{2}}(x)$ becomes$~\tan (x)$.
\Rightarrow $$$$\dfrac{d}{dx}\left( \tan x \right)={{\sec }^{2}}(x)
\Rightarrow $$$$\tan x=\int{{{\sec }^{2}}}(x)+c.
Integral is nothing but the antiderivative.
$\Rightarrow \int{{{\sec }^{^{2}}}}(x)=\tan x+c$
Here c is some constant value.
So, the antiderivative of the ${{\sec }^{2}}(x)$ becomes$~\tan (x)$.
Antiderivative of ${{\sec }^{2}}(x)$ is$~\tan (x)$+c.
Antiderivative of ${{\sec }^{2}}(x)$= $~\tan (x)$+c
\Rightarrow $$$$\dfrac{d}{dx}\left( \tan x \right)={{\sec }^{2}}(x)
$\Rightarrow$ $\int{{{\sec }^{^{2}}}}(x)=\tan (x)+c$
Antiderivative is $~\tan (x)$+c.

So, the antiderivative of ${{\sec }^{2}}(x)$ is $~\tan (x)$+c.

Note: Students must know the basis derivatives of trigonometric functions like:
$\Rightarrow \dfrac{d}{dx}\left( \tan x \right)={{\sec }^{2}}(x)$
$\Rightarrow \dfrac{d}{dx}\left( \sin x \right)=\cos x$
$\Rightarrow \dfrac{d}{dx}\left( \cos x \right)=-\sin x$
Students must know the concept of antiderivative. Students must be very careful while doing the calculations.