How does one find the integral of secant cubed of x dx?

Answer 1

Find I = ∫ sec³ x dx. The answer is I = ½ [ log(sec x + tan x) + sec x tan x ]. * Here is how we may find it: Letting s = sec x, and t = tan x, we have, s² = 1 + t², dt = s² dx = (1 + t²) dx, and ds = st dx. Then, we obtain, dI = s³ dx = s dt. * Now, d(st) = s dt + t ds = dI + t ds = dI + st² dx = dI + s(s² - 1)dx = dI + s³ dx - s dx = 2dI - s dx; whence, 2dI = s dx + d(st). * Also, we have, s = (s² + st) / (s + t), whence s dx = (s² + st) dx / (s + t) = (dt + ds) / (s + t) = d(s + t) / (s + t) = d log(s + t). This gives us, 2dI = d log(s + t) + d(st). Integrating, we easily obtain, I = ½ [ log(s + t) + st ], which is the answer we sought. * Checking that we have arrived at the correct answer, we differentiate back: d(st) / dx = (st)'= st' + ts' = s³ + st² = 2s³ - s. d log(s + t) / dx = log'(s + t) = (s + t)' / (s + t) = (st + s²) / (s + t) = s. Thus, 2I' = [ st + log(s + t) ]' = 2s³; and I' = ½ [ st + log(s + t) ]' = s³, confirming that our answer is correct.