Question

$\int {\dfrac{{{{\sec }^2}x}}{{{{(\sec x + \tan x)}^{\frac{9}{2}}}}}dx = } $ (for some arbitrary constant)
A. $\dfrac{{ - 1}}{{{{(\sec x + \tan x)}^{\frac{{11}}{2}}}}}\{ \dfrac{1}{{11}} - \dfrac{1}{7}{(\sec x + \tan x)^2}\} + C$
B. $\dfrac{1}{{{{(\sec x + \tan x)}^{\frac{{11}}{2}}}}}\{ \dfrac{1}{{11}} - \dfrac{1}{7}{(\sec x + \tan x)^2}\} + C$
C. $\dfrac{{ - 1}}{{{{(\sec x + \tan x)}^{\frac{{11}}{2}}}}}\{ \dfrac{1}{{11}} + \dfrac{1}{7}{(\sec x + \tan x)^2}\} + C$
D. $\dfrac{1}{{{{(\sec x + \tan x)}^{\frac{{11}}{2}}}}}\{ \dfrac{1}{{11}} + \dfrac{1}{7}{(\sec x + \tan x)^2}\} + C$

Answer 1

Hint: The given integral contains trigonometric terms in both the numerator and the denominator, so first convert the expression into a simpler expression using substitution, whose integral can be easily evaluated.

Complete step by step answer:
The given integral is $\int {\dfrac{{{{\sec }^2}x}}{{{{(\sec x + \tan x)}^{\frac{9}{2}}}}}dx} $
Let $\sec x + \tan x = z$
Differentiating both sides with respect to $x$ , we see that :
$\sec x\tan x + {\sec ^2}xdx = dz$
$
   \Rightarrow \sec x(\sec x + \tan x)dx = dz \\
   \Rightarrow \sec xdx = \dfrac{{dz}}{z} \\
 $
We know that according to a trigonometric property -
$
  {\sec ^2}x - {\tan ^2}x = 1 \\
   \Rightarrow (\sec x - \tan x)(\sec x + \tan x) = 1 \\
   \Rightarrow \sec x - \tan x = \dfrac{1}{z} \\
 $
On adding the obtained expressions $\sec x + \tan x$ and $\sec x - \tan x$, we get :
$
  \sec x + \tan x + \sec x - \tan x = z + \dfrac{1}{z} \\
  \sec x = \dfrac{1}{2}(z + \dfrac{1}{z}) \\
 $
Substituting the obtained values of $\sec xdx$ , $\sec x$ and $\sec x + \tan x$ in the given integral, we get :
$
  \int {\dfrac{{\dfrac{1}{2}(z + \dfrac{1}{z})\dfrac{1}{z}dz}}{{{z^{\frac{9}{2}}}}}} = \dfrac{1}{2}\int {\dfrac{{{z^2} + 1}}{{{z^{\frac{{13}}{2}}}}}dz} \\
  \int {\dfrac{{\dfrac{1}{2}(z + \dfrac{1}{z})\dfrac{1}{z}dz}}{{{z^{\frac{9}{2}}}}}} = \dfrac{1}{2}[\int {\dfrac{1}{{{z^{\frac{9}{2}}}}}dz + } \int {\dfrac{1}{{{z^{\frac{{13}}{2}}}}}dz} ] \\
 $
Now, we will apply the formula of integration of ${x^n}$ , that is, $\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}} + C} $ . Thus, we get the integral of the above simplified expression as :
$
  \int {\dfrac{{\dfrac{1}{2}(z + \dfrac{1}{z})\dfrac{1}{z}dz}}{{{z^{\frac{9}{2}}}}}} = \dfrac{1}{2}(\dfrac{{ - 2}}{{7{z^{\frac{7}{2}}}}} + \dfrac{{ - 2}}{{11{z^{\frac{{11}}{2}}}}}) + C \\
  \int {\dfrac{{\dfrac{1}{2}(z + \dfrac{1}{z})\dfrac{1}{z}dz}}{{{z^{\frac{9}{2}}}}}} = \dfrac{{ - 1}}{{{z^{\frac{7}{2}}}}}(\dfrac{1}{7} + \dfrac{1}{{11{z^2}}}) + C \\
 $
Now, put the value $z = \sec x + \tan x$ in the above equation, we get :
$
  \int {\dfrac{{{{\sec }^2}x}}{{{{(\sec x + \tan x)}^{\frac{9}{2}}}}}dx} = \dfrac{{ - 1}}{{{{(\sec x + \tan x)}^{\frac{7}{2}}}}}(\dfrac{1}{7} + \dfrac{1}{{11{{(\sec x + \tan x)}^2}}}) + C \\
   \Rightarrow \int {\dfrac{{{{\sec }^2}x}}{{{{(\sec x + \tan x)}^{\frac{9}{2}}}}}dx} = \dfrac{{ - 1}}{{{{(\sec x + \tan x)}^{\frac{{11}}{2}}}}}\{ \dfrac{1}{{11}} + \dfrac{1}{7}{(\sec x + \tan x)^2}\} + C \\
 $

The correct option is option (C).

Additional information:
The integrals that contain multiple trigonometric functions are known as trigonometric integrals. (A discipline of mathematics that deals with certain angle functions and how to use them in computations is trigonometry. In trigonometry, there are six functions of an angle that are often utilized.)

Note:
In mathematics, an integral is a numerical number equal to the area under the graph of a function for some interval (definite integral) or a new function whose derivative is the original function (indefinite integral). While solving this kind of integral problems, select the part you are going to substitute such that the expression gets simpler and contains lesser number of terms and is easy to integrate. To solve trigonometric integrals, substitute the trigonometric function by some algebraic expression.