
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
588.6k+ views
Hint: 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.
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.
Recently Updated Pages
Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Business Studies: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

There are 720 permutations of the digits 1 2 3 4 5 class 11 maths CBSE

Discuss the various forms of bacteria class 11 biology CBSE

Draw a diagram of a plant cell and label at least eight class 11 biology CBSE

State the laws of reflection of light

Explain zero factorial class 11 maths CBSE

