Question

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

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

Answer 1

Answer: (A), (C)

The correct options are
A $-(\sec x-\tan x{)}^{11/2}\{\frac{1}{11}+\frac{1}{7(\sec x-\tan x{)}^2}\}+K$
C $-\frac{1}{(\sec x+\tan x{)}^{11/2}}\{\frac{1}{11}+\frac{1}{7}(\sec x+\tan x{)}^2\}+K$

$\text{Let}\sec x+\tan x=t$
$\Rightarrow \sec x(\sec x+\tan x)dx=dt$
$\text{Also},\frac{1}{t}=\sec x-\tan x(\because {\sec }^{2}x-{\tan }^{2}x=1)$
$\Rightarrow 2\sec x=t+\frac{1}{t}$
$\Rightarrow \sec x=\frac{1}{2}(t+\frac{1}{t})\cdots \left(i\right)$
$\text{Then},\int \frac{{\sec }^{2}x}{(\sec x+\tan x{)}^{9/2}}dx$
$=\int \frac{{\sec }^{2}x}{t^{9/2}}.\frac{dt}{t\sec x}$
$=\int \frac{\sec x}{t^{11/2}}dt$
$=\frac{1}{2}\int \frac{t+\frac{1}{t}}{t^{11/2}}dt\left[\text{From}\right(i\left)\right]$
$=\frac{1}{2}\int \frac{dt}{t^{9/2}}+\frac{1}{2}\int \frac{dt}{t^{13/2}}$
$=-\frac{1}{7t^{7/2}}-\frac{1}{11t^{11/2}}+K$
$=-\frac{1}{(\sec x+\tan x{)}^{11/2}}\{\frac{1}{11}+\frac{1}{7}(\sec x+\tan x{)}^2\}+K$
$=-(\sec x-\tan x{)}^{11/2}\{\frac{1}{11}+\frac{1}{7(\sec x-\tan x{)}^2}\}+K[\because \sec x+\tan x=\frac{1}{\sec x-\tan x}]$