Question

\( \int_0^1 \cos^{-1}(x) \, dx = ? \)

Answer 1

1. Integration by Parts Formula:
The integration by parts formula is: \[ \int u \, dv = uv - \int v \, du \]

2. Choose \( u \) and \( dv \):
Let: - \( u = \cos^{-1}(x) \) - \( dv = dx \)

3. Find \( du \) and \( v \):
To differentiate \( u = \cos^{-1}(x) \), we get: - \( du = -\frac{1}{\sqrt{1 - x^2}} \, dx \) To integrate \( dv = dx \), we get: - \( v = x \)

4. Apply the Integration by Parts Formula:
Now, applying the formula for integration by parts: \[ \int \cos^{-1}(x) \, dx = x \cos^{-1}(x) - \int x \left(-\frac{1}{\sqrt{1 - x^2}}\right) \, dx \] Simplifying: \[ \int \cos^{-1}(x) \, dx = x \cos^{-1}(x) + \int \frac{x}{\sqrt{1 - x^2}} \, dx \]

5. Solve the Remaining Integral:
Now, let's solve the integral: \[ \int \frac{x}{\sqrt{1 - x^2}} \, dx \] Use the substitution: - Let \( w = 1 - x^2 \), so \( dw = -2x \, dx \), or \( -\frac{dw}{2} = x \, dx \). Substituting into the integral: \[ \int \frac{x}{\sqrt{1 - x^2}} \, dx = \int \frac{1}{\sqrt{w}} \left( -\frac{dw}{2} \right) \] Simplifying: \[ = -\frac{1}{2} \int w^{-\frac{1}{2}} \, dw \] Integrating: \[ = -\frac{1}{2} \cdot \frac{w^{\frac{1}{2}}}{\frac{1}{2}} + C = -\sqrt{w} + C \] Substituting back \( w = 1 - x^2 \): \[ = -\sqrt{1 - x^2} + C \]

6. Substitute Back into the Main Integral:
Now, substitute the result from step 5 back into the main integral: \[ \int \cos^{-1}(x) \, dx = x \cos^{-1}(x) - \sqrt{1 - x^2} + C \]

7. Evaluate the Definite Integral:
Now, we evaluate the definite integral: \[ \int_0^1 \cos^{-1}(x) \, dx \] Substituting the limits of integration: \[ \left[ x \cos^{-1}(x) - \sqrt{1 - x^2} \right]_0^1 \] * When \( x = 1 \): \[ 1 \times \cos^{-1}(1) - \sqrt{1 - 1^2} = 1 \times 0 - 0 = 0 \] * When \( x = 0 \): \[ 0 \times \cos^{-1}(0) - \sqrt{1 - 0^2} = 0 - 1 = -1 \] Therefore: \[ \int_0^1 \cos^{-1}(x) \, dx = 0 - (-1) = 1 \]

Final Answer:
\[ \int_0^1 \cos^{-1}(x) \, dx = 1 \]