Question

The integral $\int \frac{\sec ^{2} x}{(\sec x+\tan x)^{9 / 2}} d x$ equals (for some arbitrary constant $K$ )

• A. $-\frac{1}{(\sec x+\tan x)^{11 / 2}}\left\{\frac{1}{11}-\frac{1}{7}(\sec x+\tan x)^{2}\right\}+K$
• B. $\frac{1}{(\sec x+\tan x)^{11 / 2}}\left\{\frac{1}{11}-\frac{1}{7}(\sec x+\tan x)^{2}\right\}+K$
• C. $-\frac{1}{(\sec x+\tan x)^{11 / 2}}\left\{\frac{1}{11}+\frac{1}{7}(\sec x+\tan x)^{2}\right\}+K$
• D. $\frac{1}{(\sec x+\tan x)^{11 / 2}}\left\{\frac{1}{11}+\frac{1}{7}(\sec x+\tan x)^{2}\right\}+K$

Answer 1

Answer: (C)

$I=\int \frac{\sec ^{2} x}{(\sec x+\tan x)^{9 / 2}} d x$
Let $ \sec x+\tan x=t$
$\Rightarrow \sec x-\tan x=1 / t$
Now $\left(\sec x \tan x+\sec ^{2} x\right) d x=d t$
$\sec x(\sec x+\tan x) d x=d t$
$\sec x d x=\frac{d t}{t}, \frac{1}{2}\left(t+\frac{1}{t}\right)=\sec x$
$I =\frac{1}{2} \int \frac{\left(t+\frac{1}{t}\right)}{t ^{9/ 2}} \frac{d t}{t}$
$=\frac{1}{2} \int\left(t^{-9 / 2}+t^{-13 / 2}\right) d t$
$=\frac{1}{2}\left[\frac{t^{-9 / 2}+1}{-\frac{9}{2}+1}+\frac{t^{-13 / 2+1}}{-\frac{13}{2}+1}\right]$
$=\frac{1}{2}\left[\frac{t^{-7 / 2}}{-\frac{7}{2}}+\frac{t^{-11} / 2}{-\frac{11}{2}}\right]$
$=-\frac{1}{7} t^{-7 / 2}-\frac{1}{11} t^{-11 / 2}$
$=-\frac{1}{7} \frac{1}{t^{7 / 2}}-\frac{1}{11} \frac{1}{t^{11 / 2}}$
$=-\frac{1}{t^{1 / 2}}\left(\frac{1}{11}+\frac{t^{2}}{7}\right)$
$=-\frac{1}{(\sec x+\tan x)^{11 / 2}}\left\{\frac{1}{t^{11/2}}+\frac{1}{7}(\sec x+\tan x)^{2}\right\}+k$