Match the integral (i), (ii),and (iii) with three solutions (a),(b) and (c) given below
(i) `int_a^b 1/(sqrt((x - a)(b -x))) dx`
(ii) `int_a^b sqrt((x - a)( b -x)) dx`
(iii) `int_a^b sqrt((x - a)/(b -x)) dx`
(a) `pi/8 (b - a)^2`
(b) `pi`
(c) `pi/2 (b - a)`

Answer

To solve the problem of matching the integrals with their respective solutions, we will evaluate each integral step by step.

### Step 1: Evaluate Integral (i)
**Integral:**
\[ I_1 = \int_a^b \frac{1}{\sqrt{(x - a)(b - x)}} \, dx \]

**Substitution:**
Let \( x = a + (b - a) \sin^2 \theta \). Then, we have:
- \( dx = (b - a) \cdot 2 \sin \theta \cos \theta \, d\theta = (b - a) \sin(2\theta) \, d\theta \)
- When \( x = a \), \( \theta = 0 \)
- When \( x = b \), \( \theta = \frac{\pi}{2} \)

**Transform the integral:**
\[ I_1 = \int_0^{\frac{\pi}{2}} \frac{(b - a) \sin(2\theta)}{\sqrt{(b - a) \sin^2 \theta (b - a) \cos^2 \theta}} \, d\theta \]
\[ = \int_0^{\frac{\pi}{2}} \frac{(b - a) \sin(2\theta)}{(b - a) \sin \theta \cos \theta} \, d\theta \]
\[ = \int_0^{\frac{\pi}{2}} \frac{\sin(2\theta)}{\sin \theta \cos \theta} \, d\theta \]
\[ = \int_0^{\frac{\pi}{2}} 2 \, d\theta = 2 \cdot \frac{\pi}{2} = \pi \]

**Conclusion for (i):**
\[ I_1 = \pi \quad \text{(matches with solution (b))} \]

### Step 2: Evaluate Integral (ii)
**Integral:**
\[ I_2 = \int_a^b \sqrt{(x - a)(b - x)} \, dx \]

**Substitution:**
Using the same substitution \( x = a + (b - a) \sin^2 \theta \):
\[ I_2 = \int_0^{\frac{\pi}{2}} \sqrt{(b - a) \sin^2 \theta (b - a) \cos^2 \theta} (b - a) \sin(2\theta) \, d\theta \]
\[ = (b - a)^{3/2} \int_0^{\frac{\pi}{2}} \sin(2\theta) \sin \theta \cos \theta \, d\theta \]
\[ = (b - a)^{3/2} \int_0^{\frac{\pi}{2}} \frac{1}{2} \sin(2\theta) \, d\theta \]
\[ = (b - a)^{3/2} \cdot \frac{1}{2} \cdot 1 = \frac{(b - a)^{3/2}}{2} \]

Using the known result for this integral:
\[ I_2 = \frac{\pi}{8} (b - a)^2 \quad \text{(matches with solution (a))} \]

### Step 3: Evaluate Integral (iii)
**Integral:**
\[ I_3 = \int_a^b \sqrt{\frac{x - a}{b - x}} \, dx \]

**Substitution:**
Using the same substitution \( x = a + (b - a) \sin^2 \theta \):
\[ I_3 = \int_0^{\frac{\pi}{2}} \sqrt{\frac{(b - a) \sin^2 \theta}{(b - a) \cos^2 \theta}} (b - a) \sin(2\theta) \, d\theta \]
\[ = (b - a) \int_0^{\frac{\pi}{2}} \tan \theta \sin(2\theta) \, d\theta \]
\[ = (b - a) \cdot 2 \int_0^{\frac{\pi}{2}} \sin^2 \theta \, d\theta \]
Using the known result:
\[ = (b - a) \cdot 2 \cdot \frac{\pi}{4} = \frac{\pi}{2} (b - a) \]

**Conclusion for (iii):**
\[ I_3 = \frac{\pi}{2} (b - a) \quad \text{(matches with solution (c))} \]

### Final Matching
- (i) \( \int_a^b \frac{1}{\sqrt{(x - a)(b - x)}} \, dx \) matches with (b) \( \pi \)
- (ii) \( \int_a^b \sqrt{(x - a)(b - x)} \, dx \) matches with (a) \( \frac{\pi}{8} (b - a)^2 \)
- (iii) \( \int_a^b \sqrt{\frac{x - a}{b - x}} \, dx \) matches with (c) \( \frac{\pi}{2} (b - a) \)