If `I=int(dx)/sqrt((1-x)(x-2))`, then I is equal to

Options

A. `sin^(-1)(2x-3)+C`
B. `sin^(-1)(2x+5)+C`
C. `sin^(-1)(3-2x)+C`
D. `sin^(-1)(5-2x)+C`

Answer

To solve the integral \( I = \int \frac{dx}{\sqrt{(1-x)(x-2)}} \), we will follow a series of steps to simplify and evaluate the integral.

### Step-by-Step Solution:

1. **Rewrite the Integral**:
\[ I = \int \frac{dx}{\sqrt{(1-x)(x-2)}} \]

2. **Simplify the Expression Under the Square Root**:
We can express the square root in a more manageable form:
\[ (1-x)(x-2) = -x^2 + 3x - 2 \]
Thus, we rewrite the integral:
\[ I = \int \frac{dx}{\sqrt{-x^2 + 3x - 2}} \]

3. **Complete the Square**:
To simplify \(-x^2 + 3x - 2\), we complete the square:
\[ -x^2 + 3x - 2 = -\left(x^2 - 3x + \frac{9}{4} - \frac{9}{4} + 2\right) = -\left((x - \frac{3}{2})^2 - \frac{1}{4}\right) \]
Therefore:
\[ -x^2 + 3x - 2 = -\left((x - \frac{3}{2})^2 - \frac{1}{4}\right) = \frac{1}{4} - (x - \frac{3}{2})^2 \]

4. **Substitute Back into the Integral**:
Now substitute back into the integral:
\[ I = \int \frac{dx}{\sqrt{\frac{1}{4} - (x - \frac{3}{2})^2}} \]

5. **Use a Trigonometric Substitution**:
Let \( x - \frac{3}{2} = \frac{1}{2} \sin \theta \), then \( dx = \frac{1}{2} \cos \theta \, d\theta \):
\[ I = \int \frac{\frac{1}{2} \cos \theta \, d\theta}{\sqrt{\frac{1}{4} - \frac{1}{4} \sin^2 \theta}} = \int \frac{\frac{1}{2} \cos \theta \, d\theta}{\frac{1}{2} \sqrt{1 - \sin^2 \theta}} = \int \cos \theta \, d\theta \]

6. **Integrate**:
The integral of \( \cos \theta \) is:
\[ I = \sin \theta + C \]

7. **Back Substitute**:
Recall that \( \sin \theta = \frac{2(x - \frac{3}{2})}{1} = 2x - 3 \):
\[ I = 2x - 3 + C \]

### Final Answer:
Thus, the value of the integral is:
\[ I = 2x - 3 + C \]