Question

$\int \frac{{\sec }^{2}x}{{(\sec x+\tan x)}^{9/2}}dx$ equals
(where $C$ is constant of integration)

Answer 1

Let $I=\int \frac{{\sec }^{2}x}{{(\sec x+\tan x)}^{9/2}}dx$
Put $\sec x+\tan x=t$
$\Rightarrow \sec x-\tan x=\frac{1}{t}$
$\Rightarrow \sec x=\frac{1}{2}(t+\frac{1}{t})$
Also, $\sec x(\sec x+\tan x)dx=dt$
$\Rightarrow \sec xdx=\frac{dt}{t}$
$\therefore I=\frac{1}{2}\int \frac{(t+\frac{1}{t})dt}{t^{9/2}.t}$
$\Rightarrow I=\frac{1}{2}\int (t^{-9/2}+t^{-13/2})dt$
$\Rightarrow I=\frac{1}{2}[-\frac{2}{7}\cdot t^{-7/2}-\frac{2}{11}\cdot t^{-11/2}]+C$
$\Rightarrow I=-\frac{1}{7\cdot t^{7/2}}-\frac{1}{11\cdot t^{11/2}}+C$
$\Rightarrow I=-\frac{1}{t^{11/2}}(\frac{1}{11}+\frac{t^2}{7})+C$
$\therefore I=\frac{-1}{(\sec x+\tan x{)}^{11/2}}\{\frac{1}{11}+\frac{1}{7}(\sec x+\tan x{)}^2\}+C$