
Evaluate the following integrals $\int {{{\sec }^4}x\tan xdx} $
Answer
232.8k+ views
Hint: Before attempting this question, one should have prior knowledge about the concept of integration and also remember to use $1 + {\tan ^2}x = {\sec ^2}x$ and take $\tan x$ as t and use the formula $\int {{x^n}} = \dfrac{{{x^{n + 1}}}}{{n + 1}} + C$, using this information you can approach the solution.
Complete step-by-step solution:
According to the given information we have integrals as $\int {{{\sec }^4}x\tan xdx} $
We can rewrite the given integrals as $\int {{{\sec }^2}x{{\sec }^2}x\tan xdx} $
As we know that by the trigonometric identity $1 + {\tan ^2}x = {\sec ^2}x$
Therefore, $\int {\left( {1 + {{\tan }^2}x} \right){{\sec }^2}x\tan xdx} $ (equation 1)
Let $\tan x$= t
Differentiating both side with respect to x we get
\[\dfrac{d}{{dx}}\left( {\tan x} \right) = \dfrac{{dt}}{{dx}}\]
We know that \[\dfrac{d}{{dx}}\left( {\tan x} \right) = {\sec ^2}x\]
So, \[{\sec ^2}x = \dfrac{{dt}}{{dx}}\]
\[ \Rightarrow \]\[dx = \dfrac{{dt}}{{{{\sec }^2}x}}\]
Substituting the values in equation 1 we get
$\int {\left( {1 + {t^2}} \right)t \times {{\sec }^2}x \times \dfrac{{dt}}{{{{\sec }^2}x}}} $
\[ \Rightarrow \]$\int {\left( {1 + {t^2}} \right)tdt} $
\[ \Rightarrow \]$\int {\left( {t + {t^3}} \right)dt} $
\[ \Rightarrow \]$\int {tdt} + \int {{t^3}dt} $
We know that by the integration identity $\int {{x^n}} = \dfrac{{{x^{n + 1}}}}{{n + 1}} + C$
Therefore, $\int {tdt} + \int {{t^3}dt} = \dfrac{{{t^{1 + 1}}}}{{1 + 1}} + \dfrac{{{t^{3 + 1}}}}{{3 + 1}}$
\[ \Rightarrow \]$\dfrac{{{t^2}}}{2} + \dfrac{{{t^4}}}{4} + C$
Now substituting the value of t in the above equation we get
$\dfrac{{{{\tan }^2}x}}{2} + \dfrac{{{{\tan }^4}x}}{4} + C$
Therefore, the integral of $\int {{{\sec }^4}x\tan xdx} = \dfrac{{{{\tan }^2}x}}{2} + \dfrac{{{{\tan }^4}x}}{4} + C$.
Note: In the above question we used the method of integration to find the integral of trigonometric functions where the term “function” can be explained as a relation between the provided inputs and the outputs of the given inputs such that each input is directly related to the one output. The representation of a function is given by supposing if there is a function “f” that belongs from X to Y then the function is represented by $f:X \to Y$ examples of function are one-one functions, onto functions, bijective functions, trigonometric function, binary function, etc.
Complete step-by-step solution:
According to the given information we have integrals as $\int {{{\sec }^4}x\tan xdx} $
We can rewrite the given integrals as $\int {{{\sec }^2}x{{\sec }^2}x\tan xdx} $
As we know that by the trigonometric identity $1 + {\tan ^2}x = {\sec ^2}x$
Therefore, $\int {\left( {1 + {{\tan }^2}x} \right){{\sec }^2}x\tan xdx} $ (equation 1)
Let $\tan x$= t
Differentiating both side with respect to x we get
\[\dfrac{d}{{dx}}\left( {\tan x} \right) = \dfrac{{dt}}{{dx}}\]
We know that \[\dfrac{d}{{dx}}\left( {\tan x} \right) = {\sec ^2}x\]
So, \[{\sec ^2}x = \dfrac{{dt}}{{dx}}\]
\[ \Rightarrow \]\[dx = \dfrac{{dt}}{{{{\sec }^2}x}}\]
Substituting the values in equation 1 we get
$\int {\left( {1 + {t^2}} \right)t \times {{\sec }^2}x \times \dfrac{{dt}}{{{{\sec }^2}x}}} $
\[ \Rightarrow \]$\int {\left( {1 + {t^2}} \right)tdt} $
\[ \Rightarrow \]$\int {\left( {t + {t^3}} \right)dt} $
\[ \Rightarrow \]$\int {tdt} + \int {{t^3}dt} $
We know that by the integration identity $\int {{x^n}} = \dfrac{{{x^{n + 1}}}}{{n + 1}} + C$
Therefore, $\int {tdt} + \int {{t^3}dt} = \dfrac{{{t^{1 + 1}}}}{{1 + 1}} + \dfrac{{{t^{3 + 1}}}}{{3 + 1}}$
\[ \Rightarrow \]$\dfrac{{{t^2}}}{2} + \dfrac{{{t^4}}}{4} + C$
Now substituting the value of t in the above equation we get
$\dfrac{{{{\tan }^2}x}}{2} + \dfrac{{{{\tan }^4}x}}{4} + C$
Therefore, the integral of $\int {{{\sec }^4}x\tan xdx} = \dfrac{{{{\tan }^2}x}}{2} + \dfrac{{{{\tan }^4}x}}{4} + C$.
Note: In the above question we used the method of integration to find the integral of trigonometric functions where the term “function” can be explained as a relation between the provided inputs and the outputs of the given inputs such that each input is directly related to the one output. The representation of a function is given by supposing if there is a function “f” that belongs from X to Y then the function is represented by $f:X \to Y$ examples of function are one-one functions, onto functions, bijective functions, trigonometric function, binary function, etc.
Recently Updated Pages
JEE Main 2023 April 6 Shift 1 Question Paper with Answer Key

JEE Main 2023 April 6 Shift 2 Question Paper with Answer Key

JEE Main 2023 (January 31 Evening Shift) Question Paper with Solutions [PDF]

JEE Main 2023 January 30 Shift 2 Question Paper with Answer Key

JEE Main 2023 January 25 Shift 1 Question Paper with Answer Key

JEE Main 2023 January 24 Shift 2 Question Paper with Answer Key

Trending doubts
JEE Main 2026: Session 2 Registration Open, City Intimation Slip, Exam Dates, Syllabus & Eligibility

JEE Main 2026 Application Login: Direct Link, Registration, Form Fill, and Steps

Understanding the Angle of Deviation in a Prism

Hybridisation in Chemistry – Concept, Types & Applications

How to Convert a Galvanometer into an Ammeter or Voltmeter

Understanding the Electric Field of a Uniformly Charged Ring

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

Derivation of Equation of Trajectory Explained for Students

Understanding Electromagnetic Waves and Their Importance

Understanding How a Current Loop Acts as a Magnetic Dipole

Understanding Average and RMS Value in Electrical Circuits

