Final answer:
To evaluate the integral ∫ sin⁻¹(x) dx, we can use the method of integration by parts and a trigonometric substitution. The result is x sin⁻¹(x) + ln|√(1-x²)| + C.
Step-by-step explanation:
To evaluate the integral ∫ sin-1(x) dx, we can use the method of integration by parts. Let u = sin-1(x) and dv = dx. Then, du = (1/√(1-x²)) dx and v = x.
Using the formula ∫ u dv = uv - ∫ v du, we have:
∫ sin-1(x) dx = x sin-1(x) - ∫ x/√(1-x²) dx.
Now, we can evaluate the remaining integral using a trigonometric substitution. Let x = sin(θ) or θ = sin-1(x). Then, dx = cos(θ) dθ.
Substituting these values, we get:
∫ x/√(1-x²) dx = ∫ sin(θ) / cos(θ) dθ = ∫ tan(θ) dθ = -ln|cos(θ)| + C.
Finally, substituting back x = sin(θ), we have:
∫ sin-1(x) dx = x sin-1(x) + ln|cos(sin-1(x))| + C = x sin-1(x) + ln|√(1-x²)| + C.