Question

Find the integration of the following.
$\int {\dfrac{{{\text{se}}{{\text{c}}^2}x}}{{{{(\sec x + \tan x)}^{\dfrac{9}{2}}}}}} dx{\text{ equals}} \\\ {\text{A}}{\text{. }} - \dfrac{1}{{{{(\sec x + \tan x)}^{\dfrac{{11}}{2}}}}}\\{ \dfrac{1}{{11}} - \dfrac{1}{7}{(\sec x + \tan x)^2}\\} + K \\\ {\text{B}}{\text{. }}\dfrac{1}{{{{(\sec x + \tan x)}^{\dfrac{{11}}{2}}}}}\\{ \dfrac{1}{{11}} - \dfrac{1}{7}{(\sec x + \tan x)^2}\\} + K \\\ {\text{C}}{\text{. }} - \dfrac{1}{{{{(\sec x + \tan x)}^{\dfrac{{11}}{2}}}}}\\{ \dfrac{1}{{11}} + \dfrac{1}{7}{(\sec x + \tan x)^{}}\\} + K \\\ {\text{D}}{\text{. None of the above}} \\\$

Answer 1

In this question first we have to let $\sec x{\text{ + }}\tan x{\text{ = t }}$then using trigonometric relations we convert the given question into simpler form like $\int {{x^n}dx}$and then integrate it using formula $\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}}}$ to get answer in term of t then we substitute of t equal to$\sec x{\text{ + }}\tan x{\text{ }}$for getting answer in terms of $x$

Complete step-by-step answer:
${\text{Let I = }}\int {\dfrac{{{\text{se}}{{\text{c}}^2}x}}{{{{(\sec x + \tan x)}^{\dfrac{9}{2}}}}}} dx{\text{ eq}}{\text{.1}}$
Now, again let $\sec x{\text{ + }}\tan x{\text{ = t}}$ eq.2
We know, ${\text{se}}{{\text{c}}^2}x - {\text{ta}}{{\text{n}}^2}x = 1$

$$\Rightarrow {\text{ }}(\sec x + \tan x)(\sec x - \tan x) = 1{\text{ }} \\\ \Rightarrow {\text{ t}}(\sec x - \tan x) = 1{\text{ \\{ }}\therefore {\text{from eq}}{\text{.1\\} }} \\\ \Rightarrow {\text{ }}(\sec x - \tan x) = \dfrac{1}{{\text{t}}}{\text{ eq}}{\text{.3 }} \\\$$

Add eq.2 and eq.3 we get

$$\Rightarrow {\text{ }}(\sec x - \tan x) + (\sec x{\text{ + }}\tan x) = \dfrac{1}{{\text{t}}}{\text{ + t }} \\\ \Rightarrow {\text{ }}\sec x + \sec x = \dfrac{1}{{\text{t}}}{\text{ + t}} \\\ \Rightarrow {\text{ }}\sec x = \dfrac{1}{2}(\dfrac{1}{{\text{t}}}{\text{ + t) eq}}{\text{.4}} \\\$$

We know the differentiation of $\sec x{\text{ = }}\sec x\tan x$and of $\tan x = {\sec ^2}x{\text{ }}$
Then, we apply $\dfrac{{dx}}{{dt}}$ on eq.2, we get
$\Rightarrow {\text{ }}(\sec x\tan x + {\sec ^2}x)dx = dt$
On taking $\sec x$ common from above equation, we get

$$\Rightarrow {\text{ }}\sec x(\tan x + \sec x)dx = dt \\\ \Rightarrow {\text{ }}\sec xdx = \dfrac{{dt}}{{(\tan x + \sec x)}} \\\ \Rightarrow {\text{ }}\sec xdx = \dfrac{{dt}}{{\text{t}}}{\text{ \\{ }}\therefore {\text{from eq}}{\text{.2\\} eq}}{\text{.5}} \\\$$

Now with the help of eq.2, eq.3, eq.4, we can rewrite eq.1 as
$\Rightarrow {\text{ I = }}\int {\dfrac{1}{2}\dfrac{{({\text{t + }}\dfrac{1}{{\text{t}}})}}{{{{\text{t}}^{\dfrac{9}{2}}}}}\dfrac{{dt}}{{\text{t}}}}$
On simplifying the above equation
$\Rightarrow {\text{ I = }}\dfrac{1}{2}\int {({{\text{t}}^{\dfrac{{ - 9}}{2}}}{\text{ + }}{{\text{t}}^{\dfrac{{ - 13}}{2}}})dt}$
We know, $\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}}}$
Then,

$$\Rightarrow {\text{ I = }}\dfrac{1}{2}\\{ (\dfrac{{{{\text{t}}^{\dfrac{{ - 9}}{2} + 1}}}}{{^{\dfrac{{ - 9}}{2} + 1}}}{\text{) + (}}\dfrac{{{{\text{t}}^{\dfrac{{ - 13}}{2} + 1}}}}{{^{\dfrac{{ - 13}}{2} + 1}}})\\} {\text{ + }}K \\\ \Rightarrow {\text{ I = }}\dfrac{1}{2}\\{ (\dfrac{{{{\text{t}}^{\dfrac{{ - 7}}{2}}}}}{{^{\dfrac{{ - 7}}{2}}}}{\text{) + (}}\dfrac{{{{\text{t}}^{\dfrac{{ - 11}}{2}}}}}{{^{\dfrac{{ - 11}}{2}}}})\\} {\text{ + }}K \\\ \Rightarrow {\text{ I = }} - \\{ (\dfrac{{{{\text{t}}^{\dfrac{{ - 7}}{2}}}}}{7}{\text{) + (}}\dfrac{{{{\text{t}}^{\dfrac{{ - 11}}{2}}}}}{{{{11}^{}}}})\\} {\text{ + }}K \\\$$

On taking ${{\text{t}}^{\dfrac{{ - 11}}{2}}}$ common from above equation, we get
$\Rightarrow {\text{ I = }} - {{\text{t}}^{\dfrac{{ - 11}}{2}}}\\{ (\dfrac{{{{\text{t}}^2}}}{7}{\text{) + (}}\dfrac{1}{{{{11}^{}}}})\\} {\text{ + }}K$
Now, put ${\text{t = }}\sec x{\text{ + }}\tan x$ from eq.2, we get

$$\Rightarrow {\text{ I = }} - \dfrac{1}{{{{(\sec x{\text{ + }}\tan x)}^{\dfrac{{11}}{2}}}}}\\{ \dfrac{{{{(\sec x{\text{ + }}\tan x)}^2}}}{7}{\text{ + }}\dfrac{1}{{{{11}^{}}}}\\} {\text{ + }}K \\\ \Rightarrow \int {\dfrac{{{\text{se}}{{\text{c}}^2}x}}{{{{(\sec x + \tan x)}^{\dfrac{9}{2}}}}}} dx{\text{ = }} - \dfrac{1}{{{{(\sec x{\text{ + }}\tan x)}^{\dfrac{{11}}{2}}}}}\\{ \dfrac{{{{(\sec x{\text{ + }}\tan x)}^2}}}{7}{\text{ + }}\dfrac{1}{{{{11}^{}}}}\\} {\text{ + }}K \\\$$

So, the correct answer is “Option D”.

Note: Whenever you get this type of problem the key concept of solving is to observe the trigonometric functions involved in it. And what to let as in this question we letting the $\sec x{\text{ + }}\tan x{\text{ }}$= t to convert expression into some simpler expression of which we know the integration ($\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}}}$). And one more thing, at least on obtaining an answer in terms of letting variables you have to put its value that you let at the starting of the question.