Evaluate : `intdx/sqrt((x-1)(2-x))` by the substitution `x=1+sin^2theta`. Hence, find the value of `int_1^2dx/sqrt((x-1)(2-x))`

Answer

To evaluate the integral \( I = \int \frac{dx}{\sqrt{(x-1)(2-x)}} \) using the substitution \( x = 1 + \sin^2 \theta \), we will follow these steps:

### Step 1: Substitution
Let \( x = 1 + \sin^2 \theta \).

### Step 2: Differentiate
Differentiating both sides, we get:
\[ dx = 2 \sin \theta \cos \theta \, d\theta = \sin(2\theta) \, d\theta \]

### Step 3: Substitute in the Integral
Now, substitute \( x \) and \( dx \) into the integral:
\[ I = \int \frac{dx}{\sqrt{(x-1)(2-x)}} = \int \frac{2 \sin \theta \cos \theta \, d\theta}{\sqrt{(1 + \sin^2 \theta - 1)(2 - (1 + \sin^2 \theta))}} \]

### Step 4: Simplify the Expression
Now simplify the expression inside the square root:
\[ (x-1) = \sin^2 \theta \]
\[ (2-x) = 1 - \sin^2 \theta = \cos^2 \theta \]
Thus, we have:
\[ \sqrt{(x-1)(2-x)} = \sqrt{\sin^2 \theta \cos^2 \theta} = \sin \theta \cos \theta \]

### Step 5: Substitute Back
Substituting this back into the integral:
\[ I = \int \frac{2 \sin \theta \cos \theta \, d\theta}{\sin \theta \cos \theta} = \int 2 \, d\theta \]

### Step 6: Integrate
Now we can integrate:
\[ I = 2\theta + C \]

### Step 7: Back Substitute
Since \( \sin^2 \theta = x - 1 \), we have:
\[ \theta = \sin^{-1}(\sqrt{x - 1}) \]
Thus,
\[ I = 2 \sin^{-1}(\sqrt{x - 1}) + C \]

### Step 8: Evaluate the Definite Integral
Now, we need to evaluate the definite integral from 1 to 2:
\[ \int_1^2 \frac{dx}{\sqrt{(x-1)(2-x)}} = \left[ 2 \sin^{-1}(\sqrt{x - 1}) \right]_1^2 \]

### Step 9: Calculate the Limits
Calculating the upper limit:
\[ \text{At } x = 2: \quad 2 \sin^{-1}(\sqrt{2 - 1}) = 2 \sin^{-1}(1) = 2 \cdot \frac{\pi}{2} = \pi \]
Calculating the lower limit:
\[ \text{At } x = 1: \quad 2 \sin^{-1}(\sqrt{1 - 1}) = 2 \sin^{-1}(0) = 0 \]

### Step 10: Final Result
Thus, the value of the definite integral is:
\[ \int_1^2 \frac{dx}{\sqrt{(x-1)(2-x)}} = \pi - 0 = \pi \]

### Summary
The final answer is:
\[ \int_1^2 \frac{dx}{\sqrt{(x-1)(2-x)}} = \pi \]