Question

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

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

Answer 1

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

Put $\sec x+\tan x=t$
$\sec x(\sec x+\tan x)dx=dt$
$\sec xdx=\frac{dt}{t}$
and $\sec x-\tan x=\frac{1}{t}$
$\sec x=\frac{t+\frac{1}{t}}{2}$
$\int \frac{\sec x}{t.t^{9/2}}dt=\int \frac{1}{2}\frac{(t+\frac{1}{t})}{t.t^{9/2}}dt$
$=\frac{1}{2}\int (\frac{1}{t^{9/2}}+\frac{1}{t^{13/2}})dt$
$=-\frac{1}{2}[\frac{2}{7t^{7/2}}+\frac{1}{11t^{11/2}}]+k$
$=-\frac{1}{t^{11/2}}[\frac{t^2}{7}+\frac{1}{11}]+k$
$=\frac{-1}{(\sec x+\tan x{)}^{\frac{11}{2}}}\{\frac{1}{11}+\frac{1}{7}(\sec x+\tan x{)}^2\}+k$