Question

∫01cos-1x dx = ?

Answer 1

1. Integration by Parts Formula:
The integration by parts formula is: ∫ u dv = uv - ∫ v du

2. Choose u and dv:
* Let u = cos⁻¹(x)
* Let dv = dx

3. Find du and v:
* du = -1 / √(1 - x²) dx
* v = x

4. Apply the Integration by Parts Formula:
∫ cos⁻¹(x) dx = x cos⁻¹(x) - ∫ x (-1 / √(1 - x²)) dx
∫ cos⁻¹(x) dx = x cos⁻¹(x) + ∫ x / √(1 - x²) dx

5. Solve the Remaining Integral:
Let's solve ∫ x / √(1 - x²) dx separately.
Use the substitution:
w = 1 - x²
dw = -2x dx
-dw/2 = x dx

∫ x / √(1 - x²) dx = ∫ 1 / √w (-dw/2)
= -1/2 ∫ w^(-1/2) dw
= -1/2 * (w^(1/2) / (1/2)) + C
= -√w + C
= -√(1 - x²) + C

6. Substitute Back into the Main Integral:
∫ cos⁻¹(x) dx = x cos⁻¹(x) - √(1 - x²) + C

7. Evaluate the Definite Integral:
∫₀¹ cos⁻¹(x) dx = [x cos⁻¹(x) - √(1 - x²)]₀¹

* When x = 1:
1 * cos⁻¹(1) - √(1 - 1²) = 1 * 0 - 0 = 0

* When x = 0:
0 * cos⁻¹(0) - √(1 - 0²) = 0 - 1 = -1

Therefore:

∫₀¹ cos⁻¹(x) dx = 0 - (-1) = 1

Final Answer:
∫₀¹ cos⁻¹(x) dx = 1