Answer:
Option C
Explanation:
Concept involved Integration by Substitution
i.e, $I=\int f(g(x)).g'(x) dx $
put , g(x)=t $\Rightarrow$ g'(x)dx=dt$
$\therefore$ $l=\int f(t) dt$
Description of Situation Generally,
students gets confused after substitution i.e.,
$\sec x+\tan x=t$
Now, for sec x we should use
$\sec^{2}x-tan^{2}x=1$
$\Rightarrow$ $(\sec x-\tan x)(\sec x+\tan x)=1$
$\Rightarrow$ $\sec x-\tan x=\frac{1}{t}$
sol. Here, $\int \frac{\sec^{2}x}{(\sec x+\tan x)^{9/2}}dx$
Put, $\sec x+\tan x=t$
$\Rightarrow$ $(\sec x \tan x+\sec^{2}x) dx=dt$
$\Rightarrow$ $\sec x .t dx=dt$
$\Rightarrow$ $\sec x dx= \frac{dt}{t}$
$\therefore$ $\sec x-\tan x= \frac{1}{t}$
$\Rightarrow$ $\sec x=\frac{1}{2}\left(t+\frac{1}{t}\right)$
$\therefore$ $l=\int \frac{\sec x.\sec x dx}{(\sec x+\tan x)^{9/2}}$
$\Rightarrow$ $l=\int \frac{\frac{1}{2}\left(t+\frac{1}{t}\right).\frac{dt}{t}}{t^{9/2}}$
$=\frac{1}{2}\int\left(\frac{1}{t^{9/2}}+\frac{1}{t^{13/2}}\right)dt$
=$-\frac{1}{2}\left\{\frac{2}{7 t^{7/2}}+\frac{2}{11t^{11/2}}\right\}+K$
=$-\left[\frac{1}{7(\sec x+\tan x)^{1/2}}+\frac{1}{11(\sec x+\tan x)^{11/2}}\right]+K$
$=\frac{-1}{(\sec x+\tan x)^{11/2}}$
$\left\{\frac{1}{11}+\frac{1}{7}(\sec x+\tan x)^{2}\right\}+K$
