Question

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

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

Answer 1

The correct option is C $-\frac{1}{(\sec x+\tan x{)}^{11/2}}\{\frac{1}{11}+\frac{1}{7}(\sec x+\tan x{)}^2\}+k$

$I=\int \frac{{\sec }^{2}x}{(\sec x+\tan x{)}^{9/2}}dx$
Let $\sec x+\tan x=t$
or $\sec x-\tan x=1/t$
Now, $(\sec x\tan x+{\sec }^{2}x)dx=dt$
or $\sec xdx=\frac{dt}{t}$
Also, $\frac{1}{2}(t+\frac{1}{t})=\sec x$
$\therefore I=\frac{1}{2}\int \frac{(t+\frac{1}{t})}{t^{9/2}}\frac{dt}{t}$
$=\frac{1}{2}\int (t^{-9/2}+t^{-13/2})dt$

$=\frac{1}{2}[\frac{t^{-\frac{9}{2}+1}}{\frac{-9}{2}+1}+\frac{t^{-\frac{13}{2}+1}}{-\frac{13}{2}+1}]+k$

$=\frac{1}{2}[\frac{t^{-7/2}}{\frac{-7}{2}}+\frac{t^{-11/2}}{-\frac{11}{2}}]+k$

$=-\frac{1}{7}{t}^{-7/2}-\frac{1}{11}{t}^{-11/2}+k$

$=-\frac{1}{7}\frac{1}{t^{7/2}}-\frac{1}{11}\frac{1}{t^{11/2}}+k$
$=-\frac{1}{t^{11/2}}(\frac{1}{11}+\frac{t^2}{7})$
$=-\frac{1}{(\sec x+\tan x{)}^{11/2}}\{\frac{1}{11}+\frac{1}{7}(\sec x+\tan x{)}^2\}+k$