Evaluate the integral\[\int {\dfrac{{{{\sec }^2}x}}{{{{\left( {\sec x + \tan x} \right)}^{\dfrac{9}{2}}}}}dx} \]? (For some arbitrary constant k)
$
  (a)\dfrac{{ - 1}}{{{{\left( {\sec x + \tan x} \right)}^{\dfrac{{11}}{2}}}}}\left( {\dfrac{1}{{11}} + \dfrac{{{{\left( {\sec x + \tan x} \right)}^3}}}{7}} \right) + k \\
  (b)\dfrac{{ - 1}}{{{{\left( {\sec x + \tan x} \right)}^{\dfrac{{11}}{2}}}}}\left( {\dfrac{1}{{11}} + \dfrac{{{{\left( {\sec x - \tan x} \right)}^2}}}{7}} \right) + k \\
  (c)\dfrac{{ - 1}}{{{{\left( {\sec x + \tan x} \right)}^{\dfrac{{11}}{2}}}}}\left( {\dfrac{1}{{11}} + \dfrac{{{{\left( {\sec x + \tan x} \right)}^2}}}{7}} \right) + k \\
  (d)\dfrac{1}{{{{\left( {\sec x + \tan x} \right)}^{\dfrac{{11}}{2}}}}}\left( {\dfrac{1}{{11}} + \dfrac{{{{\left( {\sec x + \tan x} \right)}^2}}}{7}} \right) + k \\
 $

Hint: In this question use the concept of substitution, let \[\sec x + \tan x\] be equal to some variable, then differentiate both sides and solve further by substituting it back to the main integral and use basic trigonometric identities to get the answer.

Complete step-by-step answer:

Let \[I = \int {\dfrac{{{{\sec }^2}x}}{{{{\left( {\sec x + \tan x} \right)}^{\dfrac{9}{2}}}}}dx} \]
Substitute \[\sec x + \tan x = t\]………………… (1)
So differentiate it w.r.t. x we have,
$ \Rightarrow \left( {\sec x\tan x + {{\sec }^2}x} \right)dx = dt$
Now simplify it we have,
$ \Rightarrow \sec x\left( {\tan x + \sec x} \right)dx = dt$
So from equation (1) we have
$ \Rightarrow \sec xdx = \dfrac{{dt}}{t}$
Now from equation (1) the value of $\dfrac{1}{t}$ =$\dfrac{1}{{\sec x + \tan x}}$………………….. (2)
Now add equation (1) and (2) we have,
$t + \dfrac{1}{t} = \sec x + \tan x + \dfrac{1}{{\sec x + \tan x}}$
Now in extreme left most part multiply and divide by $\left( {\sec x - \tan x} \right)$ we have,
\[t + \dfrac{1}{t} = \sec x + \tan x + \dfrac{1}{{\sec x + \tan x}} \times \dfrac{{\sec x - \tan x}}{{\sec x - \tan x}}\]
Now in denominator it is in the form of $\left( {a - b} \right)\left( {a + b} \right) = {a^2} - {b^2}$
\[ \Rightarrow t + \dfrac{1}{t} = \sec x + \tan x + \dfrac{{\sec x - \tan x}}{{{{\sec }^2}x - {{\tan }^2}x}}\]
And we know that the value of \[{\sec ^2}x - {\tan ^2}x\] is 1.
\[ \Rightarrow t + \dfrac{1}{t} = \sec x + \tan x + \sec x - \tan x = 2\sec x\]
$ \Rightarrow \sec x = \dfrac{1}{2}\left( {t + \dfrac{1}{t}} \right)$………………….. (3)
So from equation (1), (2) and (3) the integral becomes
\[ \Rightarrow I = \int {\dfrac{{{{\sec }^2}x}}{{{{\left( {\sec x + \tan x} \right)}^{\dfrac{9}{2}}}}}dx} = \int {\dfrac{{\sec x}}{{{{\left( t \right)}^{\dfrac{9}{2}}}}}\left( {\dfrac{{dt}}{t}} \right)} = \int {\dfrac{{\dfrac{1}{2}\left( {t + \dfrac{1}{t}} \right)}}{{{{\left( t \right)}^{\dfrac{9}{2} + 1}}}}} dt\]
Now simplify the above integral we have,
\[ \Rightarrow I = \dfrac{1}{2}\int {\dfrac{{\left( {t + \dfrac{1}{t}} \right)}}{{{{\left( t \right)}^{\dfrac{9}{2} + 1}}}}} dt = \dfrac{1}{2}\int {\left( {\dfrac{t}{{{t^{\dfrac{{11}}{2}}}}} + \dfrac{1}{{{t^{\dfrac{{11}}{2} + 1}}}}} \right)} dt = \dfrac{1}{2}\int {\left( {{t^{\dfrac{{ - 9}}{2}}} + {t^{\dfrac{{ - 13}}{2}}}} \right)dt} \]
Now as we know \[\int {{t^n}dt = \dfrac{{{t^{n + 1}}}}{{n + 1}} + k} \] where k is some arbitrary integration constant, so use this property we have,
\[ \Rightarrow I = \dfrac{1}{2}\int {\left( {{t^{\dfrac{{ - 9}}{2}}} + {t^{\dfrac{{ - 13}}{2}}}} \right)dt} = \dfrac{1}{2}\left( {\dfrac{{{t^{\dfrac{{ - 9}}{2} + 1}}}}{{\dfrac{{ - 9}}{2} + 1}} + \dfrac{{{t^{\dfrac{{ - 13}}{2} + 1}}}}{{\dfrac{{ - 13}}{2} + 1}}} \right) + k\]
Now simplify it we have,
\[ \Rightarrow I = \dfrac{1}{2}\left( { - \dfrac{2}{7}{t^{\dfrac{{ - 7}}{2}}} - \dfrac{2}{{11}}{t^{\dfrac{{ - 11}}{2}}}} \right) + k = - \dfrac{1}{7}{t^{\dfrac{{ - 7}}{2}}} - \dfrac{1}{{11}}{t^{\dfrac{{ - 11}}{2}}} + k = - \left( {\dfrac{1}{{7{t^{\dfrac{7}{2}}}}} + \dfrac{1}{{11{t^{\dfrac{{11}}{2}}}}}} \right) + k\]
\[ \Rightarrow I = \dfrac{{ - 1}}{{{t^{\dfrac{{11}}{2}}}}}\left( {\dfrac{1}{{11}} + \dfrac{{{t^2}}}{7}} \right) + k\]
Now re-substitute the value of t we have,
\[ \Rightarrow I = - \dfrac{1}{{{{\left( {\sec x + \tan x} \right)}^{\dfrac{{11}}{2}}}}}\left( {\dfrac{1}{{11}} + \dfrac{{{{\left( {\sec x + \tan x} \right)}^2}}}{7}} \right) + k\]
So this is the required value of the integral.
Hence option (c) is correct.

Note: Whenever we face such type of problems the key concept is to have good list of the basic integration formula like\[\int {{t^n}dt = \dfrac{{{t^{n + 1}}}}{{n + 1}} + k} \]. It helps in evaluating the main integral in the last after simplification. Never forget to add the constant of integration after performing the integration.