Question

Integrate the following:$\dfrac{\sqrt{{{\sin }^{4}}x+{{\cos }^{4}}x}}{{{\sin }^{3}}x\cos x}dx,x\in \left( 0,\dfrac{\pi }{2} \right)$

Answer 1

Hint: For solving this problem, first convert the whole quantity into tan function by using division. Now, substitute tan x to some other variable u. Again, substitute square of u to some other variable t. At last, substitute t as tan function of theta. Now, simplify all the variables and integrate using standard formulas.

Complete step-by-step answer:
First we have to convert whole quantity into tan function,
Dividing both the numerator and denominator by ${{\cos }^{2}}x$ and manipulating to convert whole quantity in tan, we get:
\[\begin{align}
  & \Rightarrow \dfrac{\sqrt{\dfrac{{{\sin }^{4}}x}{{{\cos }^{4}}x}+\dfrac{{{\cos }^{4}}x}{{{\cos }^{4}}x}}}{\dfrac{{{\sin }^{3}}x\cos x}{{{\cos }^{2}}x}} \\
 & \Rightarrow \dfrac{\sqrt{{{\tan }^{4}}x+1}}{{{\tan }^{2}}x\dfrac{\sin x}{\cos x}{{\cos }^{2}}x}dx \\
 & \Rightarrow \dfrac{\sqrt{{{\tan }^{4}}x+1}}{{{\tan }^{3}}x}{{\sec }^{2}}xdx \\
\end{align}\]
Let, u = tan x
Differentiation with respect to the x will give us:
$\begin{align}
  & \dfrac{du}{dx}={{\sec }^{2}}x \\
 & du={{\sec }^{2}}xdx \\
\end{align}$
As, $\dfrac{d}{dx}\left( \tan x \right)={{\sec }^{2}}x$
Replacing with variables of u in the expression, we get:
$\int{\dfrac{\sqrt{{{u}^{4}}+1}}{{{u}^{3}}}}du$
Again, let ${{u}^{2}}=t$
Differentiation with respect to u gives us:
$\begin{align}
  & 2u=\dfrac{dt}{du} \\
 & 2udu=dt \\
 & du=\dfrac{dt}{2u}=\dfrac{dt}{\sqrt{t}} \\
\end{align}$
As, $\dfrac{d}{du}\left( {{u}^{2}} \right)=2u$
Replacing the value of ${{u}^{2}}\text{ and }du$ in variables of t, we get:
$\begin{align}
  & \Rightarrow \int{\dfrac{\sqrt{{{t}^{2}}+1}}{{{t}^{\dfrac{3}{2}}}}}\times \dfrac{1}{2\sqrt{t}}dt \\
 & \Rightarrow \dfrac{1}{2}\int{\dfrac{\sqrt{{{t}^{2}}+1}}{{{t}^{2}}}}dt \\
\end{align}$
Finally, let $t=\tan \theta $
Differentiation with respect to the $\theta $ will give us:
$\begin{align}
  & \dfrac{dt}{d\theta }={{\sec }^{2}}\theta \\
 & \Rightarrow dt={{\sec }^{2}}\theta d\theta \\
\end{align}$
As, $\dfrac{d}{d\theta }\left( \tan \theta \right)={{\sec }^{2}}\theta $
Putting the value of the dt and t in the expression, we get:
$\Rightarrow \dfrac{1}{2}\int{\dfrac{\sqrt{{{\tan }^{2}}\theta +1}}{{{\tan }^{2}}\theta }}{{\sec }^{2}}\theta d\theta $
Using the identity ${{\sec }^{2}}\theta =1+{{\tan }^{2}}\theta $ and ${\dfrac{1}{{\tan}^{2}\theta}={\cot}^{2}\theta}$ , we get:
$\begin{align}
  & \Rightarrow \dfrac{1}{2}\int{\dfrac{\sec \theta }{{{\tan }^{2}}\theta }\left( {{\tan }^{2}}\theta +1 \right)d\theta } \\
 & \Rightarrow \dfrac{1}{2}\int{\sec \theta \left( \dfrac{{{\tan }^{2}}\theta +1}{{{\tan }^{2}}\theta } \right)d\theta } \\
 & \Rightarrow \dfrac{1}{2}\int{\sec \theta \left( 1+{{\cot }^{2}}\theta \right)d\theta } \\
\end{align}$
On opening the inner brackets, we get simplified expression as:
$\Rightarrow \dfrac{1}{2}\left[ \int{\sec \theta d\theta +}\int{\cot \theta \csc \theta d\theta } \right]$
The above expression is obtained by using the property:
$\begin{align}
  & \sec \theta \times {{\cot }^{2}}\theta =\dfrac{1}{\cos \theta }\times \dfrac{{{\cos }^{2}}\theta }{{{\sin }^{2}}\theta } \\
 & \therefore \dfrac{\cos \theta }{{{\sin }^{2}}\theta }=\cot \theta \csc \theta \\
\end{align}$
Integrating the above expression, we get:
$\Rightarrow \dfrac{1}{2}\left[ \int{\sec \theta d\theta +}\int{\cot \theta \csc \theta d\theta } \right]$
Now, by using standard integration results:
$\begin{align}
  & \int{\sec \theta =\ln \left| \sec \theta +\tan \theta \right|} \\
 & \int{\cot \theta \csc \theta =\csc \theta } \\
\end{align}$
$\Rightarrow \dfrac{1}{2}\left[ \ln \left| \sec \theta +\tan \theta \right|-\csc \theta \right]+c$
Replacing the value $\sec \theta ,\tan \theta \text{ and }\csc \theta $ in terms of previous substitution of t, we get:
\[\begin{align}
  & \text{As, t}=\tan \theta \\
 & \sec \theta =\sqrt{1+{{t}^{2}}},\csc \theta =\dfrac{\sqrt{1+{{t}^{2}}}}{t} \\
 & \Rightarrow \dfrac{1}{2}\left[ \ln \left| \sqrt{1+{{t}^{2}}}+t \right|-\dfrac{\sqrt{1+{{t}^{2}}}}{{{t}^{2}}} \right]+c \\
\end{align}\]
Now, replacing t with u, we get
\[\begin{align}
  & \text{As, }t={{u}^{2}} \\
 & \Rightarrow \dfrac{1}{2}\left[ \ln \left| \sqrt{1+{{u}^{4}}}+{{u}^{2}} \right|-\dfrac{\sqrt{1+{{u}^{4}}}}{{{u}^{2}}} \right]+c \\
\end{align}\]
Putting the value of u = tan x, we get:
\[\Rightarrow \dfrac{1}{2}\left[ \ln \left| \sqrt{1+{{\tan }^{4}}x}+{{\tan }^{2}}x \right|-\dfrac{\sqrt{1+{{\tan }^{4}}x}}{{{\tan }^{2}}x} \right]+c\]
The above expression is in terms of variables present in question and hence it is the final answer.

Note: Students must be careful while doing the substitution of one function to another. The corresponding multiplication factor in the differentiation should be accommodated in the same step to avoid errors. This question must be remembered by students due to three different substitution steps.