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

6Put $t=\text{sec}x+\tan x⟹\text{sec}x-\tan x=\frac{1}{t}$

$⟹\sec x=\frac{t^2+1}{2t}$

$\text{sec}x(\text{sec}x+\tan x)dx=dt⟹dx=\frac{2dt}{t^2+1}$

$\int \frac{{\text{sec}}^{2}x}{(\text{sec}x+\tan x{)}^{9/2}}dx=\int \frac{(t^2+1{)}^2}{4t^{9/2+2}}\times \frac{2dt}{t^2+1}=\frac{1}{2}\int \frac{dt}{t^{9/2}}+\int \frac{dt}{t^{13/2}}=-\frac{t^{-7/2}}{7}-\frac{t^{-11/2}}{11}=-\frac{1}{(\text{sec}x+\tan x{)}^{11/2}}(\frac{1}{11}+\frac{1}{7}(\text{sec}x+\tan x{)}^2)+K$