Question

Evaluate the value of the integral $\int{\dfrac{1}{{{\cos }^{4}}x+{{\sin }^{4}}x}dx}$.

Answer 1

First, before proceeding for this, we must know that trigonometric functions integral can be solved only when we know some of the trigonometric identities. Then, by dividing the numerator and denominator by ${{\cos }^{4}}x$and using the identities as $\dfrac{1}{\cos x}=\sec x$, $\dfrac{\sin x}{\cos x}=\tan x$and ${{\sec }^{2}}x=1+{{\tan }^{2}}x$, we get the simple integral form. Then, by using the substitution method and the formula for the integral as $\int{\dfrac{1}{{{x}^{2}}+{{a}^{2}}}dx}=\dfrac{1}{a}{{\tan }^{-1}}\dfrac{x}{a}$, we get the final result.

Complete step-by-step answer :
In this question, we are supposed to find the value of the integral $\int{\dfrac{1}{{{\cos }^{4}}x+{{\sin }^{4}}x}dx}$.
So, before proceeding for this, we must know that trigonometric functions integral can be solved only when we know some of the trigonometric identities.
Now, by dividing the numerator and denominator by ${{\cos }^{4}}x$, we get:
$\int{\dfrac{\dfrac{1}{{{\cos }^{4}}x}}{\dfrac{{{\cos }^{4}}x}{{{\cos }^{4}}x}+\dfrac{{{\sin }^{4}}x}{{{\cos }^{4}}x}}dx}$
Now, by using the trigonometric identity as $\dfrac{1}{\cos x}=\sec x$and $\dfrac{\sin x}{\cos x}=\tan x$, we get:
$\int{\dfrac{{{\sec }^{4}}x}{1+{{\tan }^{4}}x}dx}$
Now, by using the another identity as ${{\sec }^{2}}x=1+{{\tan }^{2}}x$in the above integral, we get:
$\int{\dfrac{{{\sec }^{2}}x\left( 1+{{\tan }^{2}}x \right)}{1+{{\tan }^{4}}x}dx}$
Now, let us assume that $\tan x=t$ to solve the above integral as by differentiating the assumed equation, we get:
${{\sec }^{2}}xdx=dt$
The, by replacing all the values of the integral in form of t, we get:
$\int{\dfrac{1+{{t}^{2}}}{1+{{t}^{4}}}dt}$
Now, by dividing the numerator and denominator by ${{t}^{2}}$, we get:
$\int{\dfrac{\dfrac{1}{{{t}^{2}}}+1}{\dfrac{1}{{{t}^{2}}}+{{t}^{2}}}dt}$
Now, by using the concept of the completing the square in the denominator, we get:
$\int{\dfrac{\dfrac{1}{{{t}^{2}}}+1}{{{\left( t-\dfrac{1}{t} \right)}^{2}}+2}dt}$
Then, again by using assumption that $t-\dfrac{1}{t}=u$and solve them by using differentiation as:
$\left( 1+\dfrac{1}{{{t}^{2}}} \right)dt=du$
Now, by substituting all the values in the integral in form of u, we get:
$\int{\dfrac{1}{{{u}^{2}}+2}du}$
So, we get the value of the integral by using the formula $\int{\dfrac{1}{{{x}^{2}}+{{a}^{2}}}dx}=\dfrac{1}{a}{{\tan }^{-1}}\dfrac{x}{a}$as:
$\int{\dfrac{1}{{{u}^{2}}+{{\sqrt{2}}^{2}}}du}=\dfrac{1}{\sqrt{2}}{{\tan }^{-1}}\dfrac{u}{\sqrt{2}}+c$
Now, by substituting the value of u as assumed above, we get:
$\begin{aligned} & \dfrac{1}{\sqrt{2}}{{\tan }^{-1}}\dfrac{u}{\sqrt{2}}+c=\dfrac{1}{\sqrt{2}}{{\tan }^{-1}}\dfrac{\left( t-\dfrac{1}{t} \right)}{\sqrt{2}}+c \\\ & \Rightarrow \dfrac{1}{\sqrt{2}}{{\tan }^{-1}}\dfrac{\left( \dfrac{{{t}^{2}}-1}{t} \right)}{\sqrt{2}}+c \\\ \end{aligned}$
Then, by substituting the value of t as assumed above, we get:
$\begin{aligned} & \dfrac{1}{\sqrt{2}}{{\tan }^{-1}}\dfrac{\left( \dfrac{{{t}^{2}}-1}{t} \right)}{\sqrt{2}}+c=\dfrac{1}{\sqrt{2}}{{\tan }^{-1}}\dfrac{\left( \dfrac{{{\tan }^{2}}x-1}{\tan x} \right)}{\sqrt{2}}+c \\\ & \Rightarrow \dfrac{1}{\sqrt{2}}{{\tan }^{-1}}\left( \dfrac{1}{\sqrt{2}}\left( \dfrac{{{\tan }^{2}}x-1}{\tan x} \right) \right)+c \\\ \end{aligned}$
So, we get the final value of the integral as $\dfrac{1}{\sqrt{2}}{{\tan }^{-1}}\left( \dfrac{1}{\sqrt{2}}\left( \dfrac{{{\tan }^{2}}x-1}{\tan x} \right) \right)+c$.

Note : Now, to solve these types of the questions we need to know some of the basic formulas of the differentiation used above to get the result easily. So, some of the formulas are:
$\begin{aligned} & \dfrac{d}{dx}\left( \tan x \right)={{\sec }^{2}}x \\\ & \dfrac{d}{dx}\left( \dfrac{-1}{t} \right)=\dfrac{1}{{{t}^{2}}} \\\ \end{aligned}$.