Question

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

Answer 1

Hint: 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.