Let's solve the integrals step by step.
### (i) \( \int x \sin^{-1} x \, dx \)
1. **Integration by Parts**: We will use the integration by parts formula:
\[
\int u \, dv = uv - \int v \, du
\]
Let \( u = \sin^{-1} x \) and \( dv = x \, dx \).
Then, we find \( du \) and \( v \):
\[
du = \frac{1}{\sqrt{1 - x^2}} \, dx, \quad v = \frac{x^2}{2}
\]
2. **Apply the formula**:
\[
\int x \sin^{-1} x \, dx = \sin^{-1} x \cdot \frac{x^2}{2} - \int \frac{x^2}{2} \cdot \frac{1}{\sqrt{1 - x^2}} \, dx
\]
Simplifying gives:
\[
= \frac{x^2}{2} \sin^{-1} x - \frac{1}{2} \int \frac{x^2}{\sqrt{1 - x^2}} \, dx
\]
3. **Simplify the integral**:
To simplify \( \int \frac{x^2}{\sqrt{1 - x^2}} \, dx \), we can rewrite it:
\[
\int \frac{x^2}{\sqrt{1 - x^2}} \, dx = \int \frac{1 - (1 - x^2)}{\sqrt{1 - x^2}} \, dx = \int \frac{1}{\sqrt{1 - x^2}} \, dx - \int \sqrt{1 - x^2} \, dx
\]
4. **Integrate**:
- The first integral \( \int \frac{1}{\sqrt{1 - x^2}} \, dx = \sin^{-1} x \).
- The second integral \( \int \sqrt{1 - x^2} \, dx \) can be solved using the substitution \( x = \sin \theta \).
5. **Final result**:
Combining the results gives:
\[
\int x \sin^{-1} x \, dx = \frac{x^2}{2} \sin^{-1} x - \frac{1}{2} \left( \sin^{-1} x - \frac{x}{2} \sqrt{1 - x^2} \right) + C
\]
### (ii) \( \int x \cos^{-1} x \, dx \)
1. **Integration by Parts**: Let \( u = \cos^{-1} x \) and \( dv = x \, dx \).
Then, we find \( du \) and \( v \):
\[
du = -\frac{1}{\sqrt{1 - x^2}} \, dx, \quad v = \frac{x^2}{2}
\]
2. **Apply the formula**:
\[
\int x \cos^{-1} x \, dx = \cos^{-1} x \cdot \frac{x^2}{2} + \int \frac{x^2}{2\sqrt{1 - x^2}} \, dx
\]
3. **Simplify the integral**:
Similar to the previous integral, we can rewrite:
\[
\int \frac{x^2}{\sqrt{1 - x^2}} \, dx = \int \frac{1}{\sqrt{1 - x^2}} \, dx - \int \sqrt{1 - x^2} \, dx
\]
4. **Integrate**:
- The first integral \( \int \frac{1}{\sqrt{1 - x^2}} \, dx = \sin^{-1} x \).
- The second integral \( \int \sqrt{1 - x^2} \, dx \) can be solved similarly.
5. **Final result**:
Combining gives:
\[
\int x \cos^{-1} x \, dx = \frac{x^2}{2} \cos^{-1} x + \frac{1}{2} \left( \sin^{-1} x - \frac{x}{2} \sqrt{1 - x^2} \right) + C
\]
### (iii) \( \int x \tan^{-1} x \, dx \)
1. **Integration by Parts**: Let \( u = \tan^{-1} x \) and \( dv = x \, dx \).
Then, we find \( du \) and \( v \):
\[
du = \frac{1}{1 + x^2} \, dx, \quad v = \frac{x^2}{2}
\]
2. **Apply the formula**:
\[
\int x \tan^{-1} x \, dx = \tan^{-1} x \cdot \frac{x^2}{2} - \int \frac{x^2}{2(1 + x^2)} \, dx
\]
3. **Simplify the integral**:
Rewrite the integral:
\[
\int \frac{x^2}{1 + x^2} \, dx = \int (1 - \frac{1}{1 + x^2}) \, dx = \int dx - \int \frac{1}{1 + x^2} \, dx
\]
4. **Integrate**:
- The first integral \( \int dx = x \).
- The second integral \( \int \frac{1}{1 + x^2} \, dx = \tan^{-1} x \).
5. **Final result**:
Combining gives:
\[
\int x \tan^{-1} x \, dx = \frac{x^2}{2} \tan^{-1} x - \frac{x}{2} + \frac{1}{2} \tan^{-1} x + C
\]
### (iv) \( \int x \cot^{-1} x \, dx \)
1. **Integration by Parts**: Let \( u = \cot^{-1} x \) and \( dv = x \, dx \).
Then, we find \( du \) and \( v \):
\[
du = -\frac{1}{1 + x^2} \, dx, \quad v = \frac{x^2}{2}
\]
2. **Apply the formula**:
\[
\int x \cot^{-1} x \, dx = \cot^{-1} x \cdot \frac{x^2}{2} + \int \frac{x^2}{2(1 + x^2)} \, dx
\]
3. **Simplify the integral**:
Similar to the previous integral, we can rewrite:
\[
\int \frac{x^2}{1 + x^2} \, dx = \int (1 - \frac{1}{1 + x^2}) \, dx
\]
4. **Integrate**:
- The first integral \( \int dx = x \).
- The second integral \( \int \frac{1}{1 + x^2} \, dx = \tan^{-1} x \).
5. **Final result**:
Combining gives:
\[
\int x \cot^{-1} x \, dx = \frac{x^2}{2} \cot^{-1} x + \frac{x}{2} - \frac{1}{2} \tan^{-1} x + C
\]
