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

Hint: 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.