`int_(0)^(1)cos^(-1)x dx=`

Options

A. `0`
B. 1
C. 2
D. -1

Answer

To solve the integral \( I = \int_{0}^{1} \cos^{-1}(x) \, dx \), we will use integration by parts.

### Step-by-Step Solution:

1. **Identify the parts for integration by parts**:
We will let:
\[ u = \cos^{-1}(x) \quad \text{and} \quad dv = dx \]
Then, we need to find \( du \) and \( v \):
\[ du = -\frac{1}{\sqrt{1-x^2}} \, dx \quad \text{and} \quad v = x \]

2. **Apply the integration by parts formula**:
The integration by parts formula is given by:
\[ \int u \, dv = uv - \int v \, du \]
Substituting our values, we have:
\[ I = \left[ x \cos^{-1}(x) \right]_{0}^{1} - \int_{0}^{1} x \left(-\frac{1}{\sqrt{1-x^2}}\right) \, dx \]

3. **Evaluate the boundary term**:
Now we evaluate \( \left[ x \cos^{-1}(x) \right]_{0}^{1} \):
- At \( x = 1 \):
\[ 1 \cdot \cos^{-1}(1) = 1 \cdot 0 = 0 \]
- At \( x = 0 \):
\[ 0 \cdot \cos^{-1}(0) = 0 \cdot \frac{\pi}{2} = 0 \]
Therefore,
\[ \left[ x \cos^{-1}(x) \right]_{0}^{1} = 0 - 0 = 0 \]

4. **Simplify the integral**:
Now we simplify the integral:
\[ I = 0 + \int_{0}^{1} \frac{x}{\sqrt{1-x^2}} \, dx \]

5. **Substitute for the integral**:
To evaluate \( \int_{0}^{1} \frac{x}{\sqrt{1-x^2}} \, dx \), we can use the substitution:
\[ t = 1 - x^2 \quad \Rightarrow \quad dt = -2x \, dx \quad \Rightarrow \quad dx = -\frac{dt}{2x} \]
When \( x = 0 \), \( t = 1 \) and when \( x = 1 \), \( t = 0 \). Thus:
\[ I = \int_{1}^{0} \frac{x}{\sqrt{t}} \left(-\frac{dt}{2x}\right) = \frac{1}{2} \int_{0}^{1} t^{-1/2} \, dt \]

6. **Evaluate the new integral**:
The integral \( \int t^{-1/2} \, dt \) is:
\[ \int t^{-1/2} \, dt = 2t^{1/2} \quad \text{evaluated from } 0 \text{ to } 1 \]
So we have:
\[ \frac{1}{2} \left[ 2t^{1/2} \right]_{0}^{1} = \frac{1}{2} \left[ 2(1) - 0 \right] = \frac{1}{2} \cdot 2 = 1 \]

7. **Final result**:
Therefore, the value of the integral is:
\[ I = 1 \]

### Final Answer:
\[ \int_{0}^{1} \cos^{-1}(x) \, dx = 1 \]