`(i) int_(0)^(pi//2) x cos x dx`
(i) `int_(1)^(3) x. log x dx`

Answer

Let's solve the given integrals step by step.

### Part (i): \(\int_{0}^{\frac{\pi}{2}} x \cos x \, dx\)

1. **Apply Integration by Parts**:
We will use the integration by parts formula:
\[ \int u \, dv = uv - \int v \, du \]
Let:
- \(u = x\) (which gives \(du = dx\))
- \(dv = \cos x \, dx\) (which gives \(v = \sin x\))

2. **Substitute into the formula**:
\[ \int x \cos x \, dx = x \sin x - \int \sin x \, dx \]

3. **Integrate \(\sin x\)**:
\[ \int \sin x \, dx = -\cos x \]
Therefore,
\[ \int x \cos x \, dx = x \sin x + \cos x \]

4. **Evaluate the definite integral from \(0\) to \(\frac{\pi}{2}\)**:
\[ \left[ x \sin x + \cos x \right]_{0}^{\frac{\pi}{2}} \]
- At \(x = \frac{\pi}{2}\):
\[ \frac{\pi}{2} \sin\left(\frac{\pi}{2}\right) + \cos\left(\frac{\pi}{2}\right) = \frac{\pi}{2} \cdot 1 + 0 = \frac{\pi}{2} \]
- At \(x = 0\):
\[ 0 \cdot \sin(0) + \cos(0) = 0 + 1 = 1 \]

5. **Final Calculation**:
\[ \left[ x \sin x + \cos x \right]_{0}^{\frac{\pi}{2}} = \frac{\pi}{2} - 1 \]

Thus, the answer for part (i) is:
\[ \int_{0}^{\frac{\pi}{2}} x \cos x \, dx = \frac{\pi}{2} - 1 \]

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### Part (ii): \(\int_{1}^{3} x \log x \, dx\)

1. **Apply Integration by Parts**:
Let:
- \(u = \log x\) (which gives \(du = \frac{1}{x} \, dx\))
- \(dv = x \, dx\) (which gives \(v = \frac{x^2}{2}\))

2. **Substitute into the formula**:
\[ \int x \log x \, dx = \frac{x^2}{2} \log x - \int \frac{x^2}{2} \cdot \frac{1}{x} \, dx \]
Simplifying the integral:
\[ = \frac{x^2}{2} \log x - \frac{1}{2} \int x \, dx \]

3. **Integrate \(x\)**:
\[ \int x \, dx = \frac{x^2}{2} \]
Therefore,
\[ \int x \log x \, dx = \frac{x^2}{2} \log x - \frac{1}{2} \cdot \frac{x^2}{2} = \frac{x^2}{2} \log x - \frac{x^2}{4} \]

4. **Evaluate the definite integral from \(1\) to \(3\)**:
\[ \left[ \frac{x^2}{2} \log x - \frac{x^2}{4} \right]_{1}^{3} \]
- At \(x = 3\):
\[ \frac{3^2}{2} \log 3 - \frac{3^2}{4} = \frac{9}{2} \log 3 - \frac{9}{4} \]
- At \(x = 1\):
\[ \frac{1^2}{2} \log 1 - \frac{1^2}{4} = 0 - \frac{1}{4} = -\frac{1}{4} \]

5. **Final Calculation**:
\[ \left[ \frac{x^2}{2} \log x - \frac{x^2}{4} \right]_{1}^{3} = \left( \frac{9}{2} \log 3 - \frac{9}{4} \right) - \left( -\frac{1}{4} \right) \]
\[ = \frac{9}{2} \log 3 - \frac{9}{4} + \frac{1}{4} = \frac{9}{2} \log 3 - \frac{8}{4} = \frac{9}{2} \log 3 - 2 \]

Thus, the answer for part (ii) is:
\[ \int_{1}^{3} x \log x \, dx = \frac{9}{2} \log 3 - 2 \]

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