Question

The integral $\int \frac{{\sec }^{2}x}{(\sec x+\tan x{)}^{9/2}}dx$ 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)

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