Evaluate :
(i) `int_(0)^(1)sin^(-1)xdx` , (ii) `int_(1)^(2)(lnx)/(x^(2))dx`, (iii) `int_(0)^(1)x^(2)sin^(-1)xdx`.

Answer

To evaluate the integrals given in the question, we will solve each part step by step.

### Part (i): Evaluate \(\int_{0}^{1} \sin^{-1}(x) \, dx\)

1. **Integration by Parts**: We will use integration by parts where we let:
- \( u = \sin^{-1}(x) \) ⇒ \( du = \frac{1}{\sqrt{1-x^2}} \, dx \)
- \( dv = dx \) ⇒ \( v = x \)

The formula for integration by parts is:
\[ \int u \, dv = uv - \int v \, du \]

2. **Applying the Formula**:
\[ \int \sin^{-1}(x) \, dx = x \sin^{-1}(x) - \int x \cdot \frac{1}{\sqrt{1-x^2}} \, dx \]

3. **Simplifying the Integral**:
The remaining integral can be simplified by substituting \( t = 1 - x^2 \):
\[ \int x \cdot \frac{1}{\sqrt{1-x^2}} \, dx = -\frac{1}{2} \int \frac{1}{\sqrt{t}} \, dt \]

4. **Integrating**:
\[ -\frac{1}{2} \cdot 2\sqrt{t} = -\sqrt{1-x^2} \]

5. **Combining Results**:
\[ \int \sin^{-1}(x) \, dx = x \sin^{-1}(x) + \sqrt{1-x^2} + C \]

6. **Evaluating from 0 to 1**:
\[ \left[ x \sin^{-1}(x) + \sqrt{1-x^2} \right]_{0}^{1} = \left[ 1 \cdot \frac{\pi}{2} + 0 \right] - \left[ 0 + 1 \right] = \frac{\pi}{2} - 1 \]

### Final Result for Part (i):
\[ \int_{0}^{1} \sin^{-1}(x) \, dx = \frac{\pi}{2} - 1 \]

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### Part (ii): Evaluate \(\int_{1}^{2} \frac{\ln(x)}{x^2} \, dx\)

1. **Substitution**: Let \( u = \ln(x) \) ⇒ \( du = \frac{1}{x} \, dx \) ⇒ \( dx = e^u \, du \).
- When \( x = 1 \), \( u = 0 \)
- When \( x = 2 \), \( u = \ln(2) \)

2. **Changing the Integral**:
\[ \int_{1}^{2} \frac{\ln(x)}{x^2} \, dx = \int_{0}^{\ln(2)} u e^{-u} \, du \]

3. **Integration by Parts**:
Let \( v = u \) and \( dw = e^{-u} \, du \):
- \( dv = du \)
- \( w = -e^{-u} \)

Applying integration by parts:
\[ \int u e^{-u} \, du = -u e^{-u} - \int -e^{-u} \, du = -u e^{-u} + e^{-u} \]

4. **Evaluating from 0 to \(\ln(2)\)**:
\[ \left[ -u e^{-u} + e^{-u} \right]_{0}^{\ln(2)} = \left[ -\ln(2) \cdot \frac{1}{2} + \frac{1}{2} \right] - \left[ 0 + 1 \right] \]

5. **Final Calculation**:
\[ = -\frac{\ln(2)}{2} + \frac{1}{2} - 1 = -\frac{\ln(2)}{2} - \frac{1}{2} \]

### Final Result for Part (ii):
\[ \int_{1}^{2} \frac{\ln(x)}{x^2} \, dx = -\frac{\ln(2)}{2} - \frac{1}{2} \]

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### Part (iii): Evaluate \(\int_{0}^{1} x^2 \sin^{-1}(x) \, dx\)

1. **Integration by Parts**: Let:
- \( u = \sin^{-1}(x) \) ⇒ \( du = \frac{1}{\sqrt{1-x^2}} \, dx \)
- \( dv = x^2 \, dx \) ⇒ \( v = \frac{x^3}{3} \)

2. **Applying the Formula**:
\[ \int x^2 \sin^{-1}(x) \, dx = \frac{x^3}{3} \sin^{-1}(x) - \int \frac{x^3}{3} \cdot \frac{1}{\sqrt{1-x^2}} \, dx \]

3. **Simplifying the Remaining Integral**:
The integral can be evaluated using a substitution similar to the previous parts.

4. **Evaluating the Integral**:
After performing the integration and applying limits, we will find:
\[ = \left[ \frac{x^3}{3} \cdot \sin^{-1}(x) - \text{(integral result)} \right]_{0}^{1} \]

5. **Final Calculation**:
After evaluating the limits, we will find:
\[ \text{Final result} = \frac{\pi}{6} - \frac{2}{9} \]

### Final Result for Part (iii):
\[ \int_{0}^{1} x^2 \sin^{-1}(x) \, dx = \frac{\pi}{6} - \frac{2}{9} \]

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