The integral $\int \frac{\sec ^2 x}{(\sec x+\tan x)^{9 / 2}} d x$ equals (for some arbitrary constant K)
Answer & explanation
Correct answer: option 3
We have,
$I=\int \frac{\sec ^2 x}{(\sec x+\tan x)^{9 / 2}} d x$
Let $\sec x+\tan x=t$. Then, $\sec x-\tan x=1 / t$ and, $\sec x(\sec x+\tan x) d x=d t$.
∴ $\sec x d x=\frac{1}{t} d t$ and, $\sec x=\frac{1}{2}\left(t+\frac{1}{t}\right)$
∴ $I=\frac{1}{2} \int \frac{\frac{1}{t}\left(t+\frac{1}{t}\right)}{t^{9 / 2}} d t=\frac{1}{2} \int \frac{1}{t^{9 / 2}}+\frac{1}{t^{13 / 2}} d t$
$\Rightarrow I=-\frac{1}{7 t^{7 / 2}}-\frac{1}{11 t^{11 / 2}}+k$
$\Rightarrow I=-\frac{1}{t^{11 / 2}}\left\{\frac{t^2}{7}+\frac{1}{11}\right\}+k$
$\Rightarrow I=\frac{1}{(\sec x+\tan x)^{11 / 2}}\left\{\frac{1}{11}+\frac{1}{7}(\sec x+\tan x)^2\right\}+k$