Question:

Let t = sec x + tan x. Then dt = (sec x tan x + sec^2 x) dx = sec x (tan x + sec x) dx = t sec x dx. Also, sec x - tan x = 1/t. Adding gives 2 sec x = t + 1/t, so sec x = (t^2 + 1)/(2t). Thus, dx = dt / (t sec x) = 2 dt / (t^2 + 1). The integral becomes \int (sec^2 x / t^(9/2)) dx. Substituting sec x = (t^2 + 1)/(2t), we get \int ((t^2+1)^2 / (4t^2)) * (1/t^(9/2)) * (2/(t^2+1)) dt = 1/2 \int (t^2+1)/t^(13/2) dt = 1/2 \int (t^(-9/2) + t^(-13/2)) dt = 1/2 [t^(-7/2)/(-7/2) + t^(-11/2)/(-11/2)] + K = -1/7 t^(-7/2) - 1/11 t^(-11/2) + K = -1/(t^(11/2)) [1/11 + 1/7 t^2] + K.