Integration of √cos2x/cosx

Let 0≤x≤π/2 √(cos2x/cosx) =√(1-tan²x) ----- simple trig manipulation Let tanx = siny, -π/2≤y≤π/2 √(1-tan²x) = cosy, dx = cosy/(1+sin²y)dy, y = sin⁻¹(tanx) Thus, ∫√(cos2x/cosx)dx =∫cos²y/(1+sin²y) dy =∫-1 + 2/(cos²y+2sin²y)dy ---- simple trig manipulation = -y + ∫2/(cos²y+2sin²y)dy = -y + ∫2/(cos²y+2sin²y) * (sec²y/sec²y) dy ---- times sec²y/sec²y = -y + ∫2sec²y / (1 + 2tan²y) dy = -y + ∫√2 * (√2sec²y) / (1 + (√2tany)²) dy ---- manipulate a little = -y +√2 tan⁻(√2tany) + c ---- by u sub = - sin⁻¹(tanx) + √2 tan⁻¹(√2tan(sin⁻(tanx)) + c ----- take y back in = - sin⁻¹(tanx) + √2 tan⁻¹(√2tanx / √(1-tan²x)) + c Since √(cos2x/cosx) is even and - sin⁻¹(tanx) + √2 tan⁻¹(√2sinx / √(cos2x)) odd. The other "domain", π/2≤x≤ π, will lead to the same answer. The answer is confirmed by taking derivatives of - sin⁻¹(tanx) + √2 tan⁻¹(√2tanx / √(1-tan²x)) + c Some trig manipulations will result in the same solution as Wolfram|Alpha's: - sin⁻¹(tanx) = - tan¹(sinx/√cos2x) √2 tan⁻¹(√2tanx / √(1-tan²x)) = √2sin⁻¹(√2sinx)