Question

The integral $\int_{\dfrac{\pi }{4}}^{\dfrac{\pi }{6}} {\dfrac{{dx}}{{\sin 2x\left( {{{\tan }^5}x + {{\cot }^5}x} \right)}}}$ equals:
A) $\dfrac{1}{{10}}\left( {\dfrac{\pi }{4} - {{\tan }^{ - 1}}\left( {\dfrac{1}{{9\sqrt 3 }}} \right)} \right)$
B) $\dfrac{1}{5}\left( {\dfrac{\pi }{4} - {{\tan }^{ - 1}}\left( {\dfrac{1}{{3\sqrt 3 }}} \right)} \right)$
C) $\dfrac{\pi }{{10}}$
D) $\dfrac{1}{{20}}{\tan ^{ - 1}}\left( {\dfrac{1}{{9\sqrt 3 }}} \right)$

Answer 1

Here, we have to find the integral for the given function. First, we will use the trigonometric identity, to simplify the integrand. Then we will use the substitution method to simplify the integrand and then we will integrate the function. Integration is the process of adding the small parts to find the whole parts.

Formula Used:
We will use the following formulae:

1. Trigonometric Identity: $\sin 2x = \dfrac{{2\tan x}}{{1 + {{\tan }^2}x}}$
2. Trigonometric Identity: $\cot x = \dfrac{1}{{\tan x}}$
3. Trigonometric Identity: $1 + {\tan ^2}x = {\sec ^2}x$
4. Differentiation formula: $\dfrac{d}{{dx}}\left( {\tan x} \right) = {\sec ^2}x$; $\dfrac{d}{{dt}}\left( {{t^5}} \right) = 5{t^4}$
5. Integration formula: $\int {\dfrac{1}{{{p^2} + 1}}dp = } {\tan ^{ - 1}}p$

Complete step by step solution:
We have to find the integral $\int_{\dfrac{\pi }{4}}^{\dfrac{\pi }{6}} {\dfrac{{dx}}{{\sin 2x\left( {{{\tan }^5}x + {{\cot }^5}x} \right)}}}$.
First, we have to simplify the integrand using the trigonometric identities.
By using the trigonometric identities $\sin 2x = \dfrac{{2\tan x}}{{1 + {{\tan }^2}x}}$ and $\cot x = \dfrac{1}{{\tan x}}$, we get
$\int_{\dfrac{\pi }{4}}^{\dfrac{\pi }{6}} {\dfrac{{dx}}{{\sin 2x\left( {{{\tan }^5}x + {{\cot }^5}x} \right)}}} = \int_{\dfrac{\pi }{4}}^{\dfrac{\pi }{6}} {\dfrac{{dx}}{{\dfrac{{2\tan x}}{{1 + {{\tan }^2}x}}\left( {{{\tan }^5}x + \dfrac{1}{{{{\tan }^5}x}}} \right)}}}$
$\Rightarrow \int_{\dfrac{\pi }{4}}^{\dfrac{\pi }{6}} {\dfrac{{dx}}{{\sin 2x\left( {{{\tan }^5}x + {{\cot }^5}x} \right)}}} = \int_{\dfrac{\pi }{4}}^{\dfrac{\pi }{6}} {\dfrac{{1 + {{\tan }^2}xdx}}{{2\tan x\left( {{{\tan }^5}x + \dfrac{1}{{{{\tan }^5}x}}} \right)}}}$
Now, by using the trigonometric identity, $1 + {\tan ^2}x = {\sec ^2}x$, we get
$\Rightarrow \int_{\dfrac{\pi }{4}}^{\dfrac{\pi }{6}} {\dfrac{{dx}}{{\sin 2x\left( {{{\tan }^5}x + {{\cot }^5}x} \right)}}} = \int_{\dfrac{\pi }{4}}^{\dfrac{\pi }{6}} {\dfrac{{{{\sec }^2}xdx}}{{2\tan x\left( {{{\tan }^5}x + \dfrac{1}{{{{\tan }^5}x}}} \right)}}}$
By the property of integration, we get
$\int_{\dfrac{\pi }{4}}^{\dfrac{\pi }{6}} {\dfrac{{{{\sec }^2}xdx}}{{2\tan x\left( {{{\tan }^5}x + \dfrac{1}{{{{\tan }^5}x}}} \right)}}} = \int_{\dfrac{\pi }{6}}^{\dfrac{\pi }{4}} {\dfrac{{{{\sec }^2}xdx}}{{2\tan x\left( {{{\tan }^5}x + \dfrac{1}{{{{\tan }^5}x}}} \right)}}}$
Now, we will use the method of substitution.
Substituting $t = \tan x$, we get the limits for $t$ as $t = \tan \left( {\dfrac{\pi }{6}} \right) = \dfrac{1}{{\sqrt 3 }}$ and $t = \tan \left( {\dfrac{\pi }{4}} \right) = 1$
Differentiating $t$ with respect to $x$, we get $dt = {\sec ^2}xdx$. Then, by using the differentiation formula, we get
$\Rightarrow \int_{\dfrac{\pi }{6}}^{\dfrac{\pi }{4}} {\dfrac{{{{\sec }^2}xdx}}{{2\tan x\left( {{{\tan }^5}x + \dfrac{1}{{{{\tan }^5}x}}} \right)}}} = \int_{\dfrac{1}{{\sqrt 3 }}}^1 {\dfrac{{dt}}{{2t\left( {{t^5} + \dfrac{1}{{{t^5}}}} \right)}}}$
Simplifying the terms, we will get
$\Rightarrow \int_{\dfrac{\pi }{6}}^{\dfrac{\pi }{4}} {\dfrac{{{{\sec }^2}xdx}}{{2\tan x\left( {{{\tan }^5}x + \dfrac{1}{{{{\tan }^5}x}}} \right)}}} = \int_{\dfrac{1}{{\sqrt 3 }}}^1 {\dfrac{{dt}}{{2t\left( {\dfrac{{{t^{10}} + 1}}{{{t^5}}}} \right)}}}$
$\Rightarrow \int_{\dfrac{\pi }{6}}^{\dfrac{\pi }{4}} {\dfrac{{{{\sec }^2}xdx}}{{2\tan x\left( {{{\tan }^5}x + \dfrac{1}{{{{\tan }^5}x}}} \right)}}} = \int_{\dfrac{1}{{\sqrt 3 }}}^1 {\dfrac{{dt}}{{2\left( {\dfrac{{{t^{10}} + 1}}{{{t^4}}}} \right)}}}$
$\Rightarrow \int_{\dfrac{\pi }{6}}^{\dfrac{\pi }{4}} {\dfrac{{{{\sec }^2}xdx}}{{2\tan x\left( {{{\tan }^5}x + \dfrac{1}{{{{\tan }^5}x}}} \right)}}} = \int_{\dfrac{1}{{\sqrt 3 }}}^1 {\dfrac{{{t^4}dt}}{{2\left( {{t^{10}} + 1} \right)}}}$
Now, we will again use the method of substitution.
Substituting $p = {t^5}$, we get the limits for $p$as $p = {1^5} = 1$ and $p = {\left( {\dfrac{1}{{\sqrt 3 }}} \right)^5} = \dfrac{1}{{{3^{\dfrac{5}{2}}}}} = {3^{\dfrac{{ - 5}}{2}}}$ .
Differentiating $t$ with respect to $x$ by using the differentiation formula, we get
$\begin{array}{l}dp = 5{t^4}dt\\\ \Rightarrow {t^4}dt = \dfrac{{dp}}{5}\end{array}$
So, the equation becomes
$\Rightarrow \int_{\dfrac{1}{{\sqrt 3 }}}^1 {\dfrac{{{t^4}dt}}{{2\left( {{t^{10}} + 1} \right)}}} = \int_{{3^{\dfrac{{ - 5}}{2}}}}^1 {\dfrac{{dp}}{{2\left( {{p^2} + 1} \right) \cdot 5}}}$
$\Rightarrow \int_{\dfrac{1}{{\sqrt 3 }}}^1 {\dfrac{{{t^4}dt}}{{2\left( {{t^{10}} + 1} \right)}}} = \dfrac{1}{{10}}\int_{{3^{\dfrac{{ - 5}}{2}}}}^1 {\dfrac{{dp}}{{\left( {{p^2} + 1} \right)}}}$
Now, by integrating using the integration formula $\int {\dfrac{1}{{{p^2} + 1}}dp = } {\tan ^{ - 1}}p$, we get
$\Rightarrow \int_{\dfrac{1}{{\sqrt 3 }}}^1 {\dfrac{{{t^4}dt}}{{2\left( {{t^{10}} + 1} \right)}}} = \dfrac{1}{{10}}\left( {{{\tan }^{ - 1}}p} \right)_{{3^{\dfrac{{ - 5}}{2}}}}^1$
Now, by substituting the limits, we get
$\Rightarrow \int_{\dfrac{1}{{\sqrt 3 }}}^1 {\dfrac{{{t^4}dt}}{{2\left( {{t^{10}} + 1} \right)}}} = \dfrac{1}{{10}}\left[ {{{\tan }^{ - 1}}\left( 1 \right) - \left( {{{\tan }^{ - 1}}{3^{\dfrac{{ - 5}}{2}}}} \right)} \right]$
$\Rightarrow \int_{\dfrac{1}{{\sqrt 3 }}}^1 {\dfrac{{{t^4}dt}}{{2\left( {{t^{10}} + 1} \right)}}} = \dfrac{1}{{10}}\left[ {{{\tan }^{ - 1}}\left( 1 \right) - \left( {{{\tan }^{ - 1}}\dfrac{1}{{{3^2}\sqrt 3 }}} \right)} \right]$
$\Rightarrow \int_{\dfrac{1}{{\sqrt 3 }}}^1 {\dfrac{{{t^4}dt}}{{2\left( {{t^{10}} + 1} \right)}}} = \dfrac{1}{{10}}\left[ {\dfrac{\pi }{4} - \left( {{{\tan }^{ - 1}}\dfrac{1}{{9\sqrt 3 }}} \right)} \right]$
Therefore, The integral $\int_{\dfrac{\pi }{4}}^{\dfrac{\pi }{6}} {\dfrac{{dx}}{{\sin 2x\left( {{{\tan }^5}x + {{\cot }^5}x} \right)}}}$ equals $\dfrac{1}{{10}}\left[ {\dfrac{\pi }{4} - \left( {{{\tan }^{ - 1}}\dfrac{1}{{9\sqrt 3 }}} \right)} \right]$.

Thus Option (A) is correct.

Note:
We should always be conscious that whenever we are using the method of substitution in integration, we should always change the upper limit and lower limit accordingly to the substitution. So we should know that limits change according to the variable substituted. Here it is important to remember the different formulas of integration and differentiation.