Question

Find the solution of the integral $\int {\dfrac{1}{{{{\sin }^4}x + {{\cos }^4}x}}} dx$?

Answer 1

Hint: We have to integrate $\dfrac{1}{{{{\sin }^4}x + {{\cos }^4}x}}$ with respect to $'x'$. Let us assume the value of $\int {\dfrac{1}{{{{\sin }^4}x + {{\cos }^4}x}}} dx$ is equal to $I$. We will simplify the integrand by dividing the numerator and denominator by ${\cos ^4}x$. We also know the tangent function is the ratio of sin function to cosine function. After simplification, we will use the substitution method and we put the final answer in terms of $x$ only.

Complete step-by-step solution:
Given $\int {\dfrac{1}{{{{\sin }^4}x + {{\cos }^4}x}}} dx$
Let $I = \int {\dfrac{1}{{{{\sin }^4}x + {{\cos }^4}x}}} dx$
On dividing numerator and denominator by ${\cos ^4}x$, we get
$ \Rightarrow I = \int {\dfrac{{\dfrac{1}{{{{\cos }^4}x}}}}{{\dfrac{{{{\sin }^4}x + {{\cos }^4}x}}{{{{\cos }^4}x}}}}} dx$
Splitting the terms in denominator, we get
$ \Rightarrow I = \int {\dfrac{{\dfrac{1}{{{{\cos }^4}x}}}}{{\dfrac{{{{\sin }^4}x}}{{{{\cos }^4}x}} + \dfrac{{{{\cos }^4}x}}{{{{\cos }^4}x}}}}} dx$
As we know, $\dfrac{{\sin x}}{{\cos x}} = \tan x$. So, $\dfrac{{{{\sin }^4}x}}{{{{\cos }^4}x}} = {\tan ^4}x$ and $\dfrac{1}{{{{\cos }^4}x}} = {\sec ^4}x$
$ \Rightarrow I = \int {\dfrac{{{{\sec }^4}x}}{{{{\tan }^4}x + 1}}} dx$
Put ${\sec ^4}x$ as ${\sec ^2}x.{\sec ^2}x$
$ \Rightarrow I = \int {\dfrac{{{{\sec }^2}x.{{\sec }^2}x}}{{{{\tan }^4}x + 1}}} dx$
Put ${\sec ^2}x = 1 + {\tan ^2}x$ (identity)
$ \Rightarrow I = \int {\left\{ {\dfrac{{1 + {{\tan }^2}x}}{{1 + {{\tan }^4}x}}} \right\}} {\sec ^2}xdx$
Now to simplify thus easily we use substitution method:
Let’s put $\tan x = t$, then differentiating we have
$ \Rightarrow {\sec ^2}xdx = dt$ (As we know $\dfrac{d}{{dx}}\left( {\tan x} \right) = {\sec ^2}x$)
Then above integral becomes
$I = \int {\dfrac{{\left( {1 + {t^2}} \right)}}{{1 + {t^4}}}} dt$
Divide numerator and denominator by ${t^2}$
$I = \int {\dfrac{{\dfrac{1}{{{t^2}}} + \dfrac{{{t^2}}}{{{t^2}}}}}{{\dfrac{1}{{{t^2}}} + \dfrac{{{t^4}}}{{{t^2}}}}}} dt$
It can also be written as:
$I = \int {\dfrac{{\dfrac{1}{{{t^2}}} + 1}}{{\dfrac{1}{{{t^2}}} + {t^2}}}} dt$
\[ \Leftrightarrow I = \int {\dfrac{{1 + \dfrac{1}{{{t^2}}}}}{{{t^2} + \dfrac{1}{{{t^2}}}}}} dt\]
To make the denominator a complete square we add and subtract $2$.
\[ \Rightarrow I = \int {\dfrac{{1 + \dfrac{1}{{{t^2}}}}}{{{t^2} + \dfrac{1}{{{t^2}}} - 2 + 2}}} dt\]
Write \[{t^2} + \dfrac{1}{{{t^2}}} - 2\] as \[{\left( {t - \dfrac{1}{t}} \right)^2}\]
\[ \Rightarrow I = \int {\dfrac{{\left( {1 + \dfrac{1}{{{t^2}}}} \right)dt}}{{{{\left( {t - \dfrac{1}{t}} \right)}^2} + {{\left( {\sqrt 2 } \right)}^2}}}} \]
Let $t - \dfrac{1}{t} = v$ , then differentiating we have
$ \Rightarrow \left( {1 + \dfrac{1}{{{t^2}}}} \right)dt = dv$
Then above integral becomes
$ \Rightarrow I = \int {\dfrac{{dv}}{{{v^2} + {{\left( {\sqrt 2 } \right)}^2}}}} $
As we know, $\int {\dfrac{{dx}}{{{x^2} + {a^2}}}} = \dfrac{1}{a}{\tan ^{ - 1}}\dfrac{x}{a} + C$
Putting $x = v$ and $a = v$ , we get
$ \Rightarrow I = \dfrac{1}{{\sqrt 2 }}{\tan ^{ - 1}}\left( {\dfrac{v}{{\sqrt 2 }}} \right) + C$
But $t - \dfrac{1}{t} = v$.
$\therefore v = \dfrac{{{t^2} - 1}}{t}$
Substitute value of $v$ in the above integral
$ \Rightarrow I = \dfrac{1}{{\sqrt 2 }}{\tan ^{ - 1}}\left( {\dfrac{{{t^2} - 1}}{{\sqrt 2 t}}} \right) + C$
But we have substituted $t = \tan x$ , then we have
$ \Rightarrow I = \dfrac{1}{{\sqrt 2 }}{\tan ^{ - 1}}\left( {\dfrac{{{{\tan }^2}x - 1}}{{\sqrt 2 \tan x}}} \right) + C$
Where $C$ is the arbitrary constant called constant of integration.
Thus, the integration of $\dfrac{1}{{{{\sin }^4}x + {{\cos }^4}x}}$ is $ = \dfrac{1}{{\sqrt 2 }}{\tan ^{ - 1}}\left( {\dfrac{{{{\tan }^2}x - 1}}{{\sqrt 2 \tan x}}} \right) + C$ , where $C$ is the integration constant.

Note: We have indefinite integral that is why we added integration. If we have a definite integral we do not add integration constant. We know the integration formulas which yield inverse trigonometric functions:
(i). $\int {\dfrac{{dx}}{{\sqrt {{a^2} - {x^2}} }}} = {\sin ^{ - 1}}\dfrac{x}{a} + C$
(ii). $\int {\dfrac{{dx}}{{{x^2} + {a^2}}}} = \dfrac{1}{a}{\tan ^{ - 1}}\dfrac{x}{a} + C$
(iii). $\int {\dfrac{{dx}}{{x\sqrt {{x^2} - {a^2}} }}} = \dfrac{1}{a}{\sec ^{ - 1}}\dfrac{x}{a} + C$
We use them according to the given problem.