Question

How do you find the antiderivative of \[{\tan ^2}\left( x \right)dx\]?

Answer 1

Hint:
The given problem is based on integrals. We can apply the identities of trigonometric and integrals to solve the problem. We are trying to simplify the problem to get the solution. The given problems based on indefinite integrals, we can apply our identities and solve the problem.

Complete Step by step Solution:
 The given problem is based on the anti-derivation or we can say that the integrals.
To find the solution, we have to use identities and apply integrals. We have given \[{\tan ^2}\left( x \right)dx\].
Apply integrals, we get:
\[\int {{{\tan }^2}xdx} \]……(A)
Using identities of \[{\tan ^2}x = \dfrac{{{{\sin }^2}x}}{{{{\cos }^2}x}}\]
Substituting the value in (A), we get:
\[\int {\dfrac{{{{\sin }^2}x}}{{{{\cos }^2}x}}} \]……(B)
Again, using identities \[{\sin ^2}x = 1 - {\cos ^2}x\] in equation (B), it becomes:
\[\int {\dfrac{{1 - {{\cos }^2}x}}{{{{\cos }^2}x}}} dx\]
Separating the numerator as
\[\dfrac{1}{{{{\cos }^2}x}},\dfrac{{{{\cos }^2}x}}{{{{\cos }^2}x}}\], we get:
\[ \Rightarrow \int {\dfrac{1}{{{{\cos }^2}x}}dx} - \int {\dfrac{{{{\cos }^2}x}}{{{{\cos }^2}x}}dx} \]
By simplifying the above term
\[\int {\dfrac{1}{{{{\cos }^2}x}}dx} - \int {1dx} \]
As the identity of \[\dfrac{1}{{{{\cos }^2}x}} = {\sec ^2}x\]
So, apply it to the above term.
\[ \Rightarrow \int {{{\sec }^2}x - \int {1dx} } \]
By using the identity of integrals:
\[\int {{{\sec }^2}x = \tan x{\text{ and }}\int {1dx} } \]
It becomes:
\[\tan x - x + c\],
Where \[c\] is constant.

Note:
The above question is based on integrals. Integrals assign number to function in a way that can describe displacement, area, volume, that arise by combining infinitesimal data. Integrals are used to calculate the centre of mass, centre of gravity and also to calculate the velocity of an object.