Question

Evaluate : $$\int {\tan }^{2}xdx$$

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 : $$\int {\tan }^{2}xdx$$
Let $$I=\int {\tan }^{2}xdx$$
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
$$I=\int {\sec }^{2}xdx-\int 1dx$$
Let $$I=I_1-I_2$$
Such that ,
$$I_1=\int {\sec }^{2}xdx$$ and $$I_2=\int 1dx$$
Now , using $$\int {\sec }^{2}x=\tan x$$ and $$\int x^{n}dx={\displaystyle \frac{x^{n+1}}{n+1}}$$ , 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 , $$\int {\tan }^{2}xdx=\tan x-x+a$$ .
So, the correct answer is “$$\int {\tan }^{2}xdx=\tan x-x+a$$”.

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 :
$$\int 1dx=x+c$$
$$\int adx=ax+c$$
$$\int x^{n}dx={\displaystyle \frac{x^{n+1}}{n+1}}$$ ; $$n\ne 1$$
$$\int \sin xdx=-\cos x+c$$
$$\int \cos xdx=\sin x+c$$
$$\int {\sec }^{2}xdx=\tan x+c$$