Question

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

Answer 1

Firstly we simplify the denominator by using the formula of $\sin 2x = \dfrac{{2\tan x}}{{1 + {{\tan }^2}x}}$, then we simplify the fraction further, and then we substitute the value of ${\tan ^5}x$ as z and we change the limits and then solve the integral to get the required answer.

Complete step-by-step answer:
$\int\limits_{\dfrac{\pi }{4}}^{\dfrac{\pi }{6}} {\dfrac{{dx}}{{\sin 2x\left( {{{\tan }^5}x + {{\cot }^5}x} \right)}}}$.
We have to apply the trigonometric formula,$\sin 2x = \dfrac{{2\tan x}}{{1 + {{\tan }^2}x}}$.
Using this formula, we get
$\int\limits_{\dfrac{\pi }{4}}^{\dfrac{\pi }{6}} {\dfrac{{dx}}{{\dfrac{{2\tan x}}{{1 + {{\tan }^2}x}}\left( {{{\tan }^5}x + {{\cot }^5}x} \right)}}}$
Now we apply the formula $({\sec ^2}x = 1 + {\tan ^2}x)$, we get,
$= \int\limits_{\dfrac{\pi }{4}}^{\dfrac{\pi }{6}} {\dfrac{{{{\sec }^2}xdx}}{{2\tan x\left( {{{\tan }^5}x + {{\cot }^5}x} \right)}}}$
$Since{\text{ }}we{\text{ }}know{\text{ }}that{\text{ }}\cot x = \dfrac{1}{{\tan x}};{\text{ }}therefore,{\text{ }}{\cot ^5}x = {\left( {\dfrac{1}{{\tan x}}} \right)^5} = \dfrac{1}{{{{\tan }^5}x}}$
$= \int\limits_{\dfrac{\pi }{4}}^{\dfrac{\pi }{6}} {\dfrac{{{{\sec }^2}xdx}}{{2\tan x\left( {{{\tan }^5}x + \dfrac{1}{{{{\tan }^5}x}}} \right)}}}$
Multiplying the numerator and denominator by ${\tan ^5}x$, we get,
$= \int\limits_{\dfrac{\pi }{4}}^{\dfrac{\pi }{6}} {\dfrac{{{{\tan }^5}x \cdot {{\sec }^2}xdx}}{{2\tan x\left( {{{\tan }^{10}}x + 1} \right)}}}$
On further simplifying we get,
$= \int\limits_{\dfrac{\pi }{4}}^{\dfrac{\pi }{6}} {\dfrac{{{{\tan }^4}x \cdot {{\sec }^2}xdx}}{{2\left( {{{\tan }^{10}}x + 1} \right)}}}$
Now, we put ${\tan ^5}x = z............(1)$
Differentiating both side of this we obtain,

$$5{\tan ^4}x \cdot {\sec ^2}xdx = dz \\\ \Rightarrow {\tan ^4}x \cdot {\sec ^2}xdx = \dfrac{{dz}}{5} \\\$$

The limit of the integration will change,
$\therefore$From Eq. (1) we can get,
For $x = \dfrac{\pi }{6}$, the value of z will be, $z = {3^{ - \dfrac{5}{2}}}\left[ \because \tan \dfrac{\pi }{6} = \dfrac{1}{{\sqrt 3 }} \\\ \therefore {\tan ^5}\dfrac{\pi }{6} = {\left( {\dfrac{1}{{{3^{\dfrac{1}{2}}}}}} \right)^5} = {3^{ - \dfrac{5}{2}}} \\\ \right]$
Again, for $x = \dfrac{\pi }{4}$, the value of z will be, $z = 1\left[ \because \tan \dfrac{\pi }{4} = 1 \\\ \therefore {\tan ^5}\dfrac{\pi }{4} = 1 \\\ \right]$
Now, we write the integration in terms of z, which is

\int\limits_1^{{3^{\dfrac{{ - 5}}{2}}}} {\dfrac{1}{{2({z^2} + 1)}} \cdot \dfrac{{dz}}{5}} \\\ = \dfrac{1}{{10}}\int\limits_1^{{3^{\dfrac{{ - 5}}{2}}}} {\dfrac{{dz}}{{({z^2} + 1)}}} \\\ $$ [ We know that we can write this for constant i.e. $$\int {5f(x)dx = 5\int {f(x)dx} } $$where f(x) is a function of x. ] Here, we apply some formula of integration, which is $$\int\limits_a^b {\dfrac{{dx}}{{{x^2} + {m^2}}}} = \left[ {\dfrac{1}{m}{{\tan }^{ - 1}}\dfrac{x}{m}} \right]_a^b$$, where a, b and m are constant. $$\therefore $$ We obtain the integration as –

\dfrac{1}{{10}}\left[ {\dfrac{1}{1}{{\tan }^{ - 1}}\dfrac{z}{1}} \right]_1^{{3^{ - \dfrac{5}{2}}}}\left[ {\because Here{\text{ }}m = 1{\text{ }}and{\text{ }}a = 1,{\text{ }}b = {3^{ - \dfrac{5}{2}}}} \right] \\
= \dfrac{1}{{10}}\left[ {{{\tan }^{ - 1}}{3^{ - \dfrac{5}{2}}} - {{\tan }^{ - 1}}1} \right] \\
= \dfrac{1}{{10}}\left( {{{\tan }^{ - 1}}\dfrac{1}{{3\sqrt 3 }} - \dfrac{\pi }{4}} \right) \\

**Note:** 1.In this problem, we use various types of formula and properties of trigonometry and integration.We have to remember all the formulas and properties. 2.Also we have to know the values of $$\tan \dfrac{\pi }{6}$$ and $$\tan \dfrac{\pi }{4}$$. 3.We solve this problem by putting $$\tan x = \dfrac{{\sin x}}{{\cos x}}$$ and $$\cot x = \dfrac{{\cos x}}{{\sin x}}$$, but this method is too difficult for us. 4.We also use the relation between tan x and cot x ; tan x and sin x ; tan x and sec x.