Question

The integration of the given function within the limits $\int\limits_0^{\dfrac{\pi }{4}} {{{\tan }^2}xdx}$ is
A) $\dfrac{\pi }{4}$
B) $1 + \dfrac{\pi }{4}$
C) $1 - \dfrac{\pi }{4}$
D) $1 - \dfrac{\pi }{2}$

Answer 1

The given question integrals have an upper and lower limit, which is known as definite integration. The definite integral is the sum of areas above minus the sum of the areas below. Definite integration of a function is applicable only when the function is real and has an upper and lower limit to it.
In the question, the given function is the trigonometric function, where ${\tan ^2}x$ does not have any pre-defined integration form; hence we have to convert it to a function that has a definite integral function.

Complete step by step answer:
$I = \int\limits_0^{\dfrac{\pi }{4}} {{{\tan }^2}xdx} - - - - (i)$
The given function is real, and it is bounded in the interval $\left[ {0,\dfrac{\pi }{4}} \right]$, and there is no constant term involved.
Since ${\tan ^2}\theta = {\sec ^2}\theta - 1$, hence we can write equation (i) as the function,
$I = \int\limits_0^{\dfrac{\pi }{4}} {\left( {{{\sec }^2}x - 1} \right)dx} - - - - (ii)$
Now bifurcate the equation (ii), we get:
$I = \int\limits_0^{\dfrac{\pi }{4}} {{{\sec }^2}xdx - \int\limits_0^{\dfrac{\pi }{4}} 1 dx} - - - - (iii)$
The integration of $\int {{{\sec }^2}\theta = \tan \theta + c}$ and for a constant term $\int {1dx = x + c}$, hence we can write equation (iii) as:

$$I = \int\limits_0^{\dfrac{\pi }{4}} {{{\sec }^2}xdx - \int\limits_0^{\dfrac{\pi }{4}} 1 dx} \\\ = \left[ {\tan x} \right]_o^{\dfrac{\pi }{4}} - \left[ x \right]_0^{\dfrac{\pi }{4}} \\\ = \left[ {\tan \left( {\dfrac{\pi }{4}} \right) - \tan 0} \right] - \left[ {\dfrac{\pi }{4} - 0} \right] \\\ = \left[ {1 - 0} \right] - \dfrac{\pi }{4} \\\ = 1 - \dfrac{\pi }{4} \\\$$

Hence the value of$\int\limits_0^{\dfrac{\pi }{4}} {{{\tan }^2}xdx} = 1 - \dfrac{\pi }{4}$
Option (3) is correct.
Important properties of trigonometric functions used:
Pythagorean identities: ${\tan ^2}\theta + 1 = {\sec ^2}\theta$
Integration: $\int {{{\sec }^2}\theta = \tan \theta + c}$

Note: Students must note that in a definite integral, the given function has to be a real function and must have a definite interval $\int\limits_a^b {f\left( x \right)dx}$, where $a$ is the lower limit and $b$ is the upper limit of the function.