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 :
$\int 1 dx = x + c$
$\int a dx = ax + c$
$\int {{x^n}dx} = \dfrac{{{x^{n + 1}}}}{{n + 1}}$ ; $n \ne 1$
$\int {\sin x} dx = - \cos x + c$
$\int {\cos x} dx = \sin x + c$
$\int {{{\sec }^2}x} dx = \tan x + c$