Question

Evaluate : \[\int {{{\tan }^2}x} dx\]

Answer 1

Hint: We have to integrate the given trigonometric function \[{\tan ^2}x\] with respect to ‘\[x\]’. We solve this using integration by using the various formulas of trigonometric functions . First we change the terms of the integration by using the formula of \[{\tan ^2}x\] in terms of \[\sec x\] and then by integrating the terms after substituting the terms we get the solution of the given integral .

Complete step-by-step answer:
Given : \[\int {{{\tan }^2}x} dx\]
Let \[I = \int {{{\tan }^2}x} dx\]
Now , We have to integrate \[I\] with respect to ‘\[x\]’
As , we know that
\[{\tan ^2}x + 1 = {\sec ^2}x\]
\[{\tan ^2}x = {\sec ^2}x - 1\]
Substituting value of \[{\tan ^2}x\] in \[I\] , we get
\[I = \int {\left( {{{\sec }^2}x - 1} \right)dx} \]
Splitting the integration terms into two integrals
\[I = \int {{{\sec }^2}x} dx - \int 1 dx\]
Let \[I = {I_1} - {I_2}\]
Such that ,
\[{I_1} = \int {{{\sec }^2}x} dx\] and \[{I_2} = \int 1 dx\]
Now , using \[\int {{{\sec }^2}x} = \tan x\] and \[\int {{x^n}dx} = \dfrac{{{x^{n + 1}}}}{{n + 1}}\] , we get
\[{I_1} = \tan x + {a_1}\]
\[{I_2} = x + {a_2}\]
Now , \[I = {I_1} - {I_2}\]
\[I = \tan x - x + {a_1} + {a_2}\]
\[I = \tan x - x + a\]
(Where \[a = {a_1} + {a_2}\])
Where ‘\[{a_1}\]’ , ‘\[{a_2}\]’ , ‘\[a\]’ are integration constant
Thus , \[\int {{{\tan }^2}x} dx = \tan x - x + a\] .
So, the correct answer is “\[\int {{{\tan }^2}x} dx = \tan x - x + a\]”.

Note: As the question was of indefinite integral that’s why we added an integral constant ‘\[a\]’ to the integration . If the question would have been of definite integral then we don’t have added the integral constant to the final answer .
The formula of integration for various trigonometric terms are given as :
\[\int 1 dx = x + c\]
\[\int a dx = ax + c\]
\[\int {{x^n}dx} = \dfrac{{{x^{n + 1}}}}{{n + 1}}\] ; \[n \ne 1\]
\[\int {\sin x} dx = - \cos x + c\]
\[\int {\cos x} dx = \sin x + c\]
\[\int {{{\sec }^2}x} dx = \tan x + c\]