Answer
64.8k+ 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
Write a composition in approximately 450 500 words class 10 english JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Arrange the sentences P Q R between S1 and S5 such class 10 english JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
What is the common property of the oxides CONO and class 10 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
What happens when dilute hydrochloric acid is added class 10 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
If four points A63B 35C4 2 and Dx3x are given in such class 10 maths JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
The area of square inscribed in a circle of diameter class 10 maths JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Other Pages
In the ground state an element has 13 electrons in class 11 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
According to classical free electron theory A There class 11 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Differentiate between homogeneous and heterogeneous class 12 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Electric field due to uniformly charged sphere class 12 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Excluding stoppages the speed of a bus is 54 kmph and class 11 maths JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
A boat takes 2 hours to go 8 km and come back to a class 11 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)