Simplifying by substitution

Algebraic expressions inside an inverse function often turn into a single trigonometric function after a substitution such as x = sin θ, x = cos θ, x = tan θ or x = sec θ. The inverse then cancels, as long as the angle stays in the principal range.

sin⁻¹(2x√(1 − x²)): put x = sin θ, so √(1 − x²) = cos θ and the inside is sin 2θ. For −1/√2 ≤ x ≤ 1/√2, 2θ lies in [−π/2, π/2] and the result is 2 sin⁻¹ x.

For 1/√2 ≤ x ≤ 1, put x = cos θ instead; the same steps give 2 cos⁻¹ x. At x = √3/2 this gives π/3, while 2 sin⁻¹(√3/2) = 2π/3 would be wrong, since sin⁻¹ never exceeds π/2.

tan⁻¹(cos x/(1 − sin x)) for −3π/2 < x < π/2: write cos x = cos²(x/2) − sin²(x/2) and 1 − sin x = (cos(x/2) − sin(x/2))². After cancelling, the inside is (1 + tan(x/2))/(1 − tan(x/2)) = tan(π/4 + x/2), so the expression equals π/4 + x/2.

cot⁻¹(1/√(x² − 1)) for x > 1: put x = sec θ with θ ∈ (0, π/2). Then √(x² − 1) = tan θ, the inside is cot θ, and the result is θ = sec⁻¹ x.

The same idea gives 3 sin⁻¹ x = sin⁻¹(3x − 4x³) for x ∈ [−1/2, 1/2] and 3 cos⁻¹ x = cos⁻¹(4x³ − 3x) for x ∈ [1/2, 1], from the triple-angle formulas.

Always check the interval: it decides which substitution keeps the final angle inside the principal range, and a different interval can change the answer.