Question

The integral $\int \frac{\sec ^{2} x}{(\sec x+\tan x)^{1 / 2}} d x$ equals (for some arbitrary constant $\left.K \right)$

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

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