Question

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

• A. $-\frac{1}{(\sec x+\tan x{)}^{11/2}}\{\frac{1}{11}-\frac{1}{7}(\sec x+\tan x{)}^2\}+K$
• B. $\frac{1}{(\sec x+\tan x{)}^{11/2}}\{\frac{1}{11}-\frac{1}{7}(\sec x+\tan x{)}^2\}+K$
• C. $-\frac{1}{(\sec x+\tan x{)}^{11/2}}\{\frac{1}{11}+\frac{1}{7}(\sec x+\tan x\left)\right\}+K$
• D. None of the above

Answer 1

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

Let $I=\int \frac{{\sec }^{2}x}{(\sec x+\tan x{)}^{9/2}}dx$

Put, $\sec x+\tan x=t$

Then, $\sec x-\tan x=\frac{1}{t}$

and $\sec x(\sec x+\tan x)dx=dt$

$\Rightarrow \sec xdx=\frac{1}{t}dt$

Also, $\sec x=\frac{1}{2}(t+\frac{1}{t})$

Therefore, $I=\frac{1}{2}\int \frac{\frac{1}{t}(t+\frac{1}{t})}{t^{9/2}}dt$

$\Rightarrow I=\frac{1}{2}\int \frac{1}{t^{9/2}}+\frac{1}{t^{13/2}}dt$

$\Rightarrow I=-\frac{1}{7t^{7/2}}-\frac{1}{11t^{11/2}}+K$

$\Rightarrow I=-\frac{1}{t^{11/2}}\{\frac{t^2}{7}+\frac{1}{11}\}+K$

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