(i) `int x^2 sin x dx`
(ii) `int x^2 sin 2x dx`

Answer

Let's solve the integrals step by step.

### (i) Integral of \( \int x^2 \sin x \, dx \)

1. **Choose \( u \) and \( dv \)**:
- Let \( u = x^2 \) and \( dv = \sin x \, dx \).
- Then, differentiate \( u \) to find \( du \):
\[ du = 2x \, dx \]
- Integrate \( dv \) to find \( v \):
\[ v = -\cos x \]

2. **Apply Integration by Parts**:
- Using the formula \( \int u \, dv = uv - \int v \, du \):
\[ \int x^2 \sin x \, dx = -x^2 \cos x - \int (-\cos x)(2x) \, dx \]
- Simplifying gives:
\[ = -x^2 \cos x + 2 \int x \cos x \, dx \]

3. **Integrate \( \int x \cos x \, dx \)**:
- Again, apply integration by parts:
- Let \( u = x \) and \( dv = \cos x \, dx \).
- Then, \( du = dx \) and \( v = \sin x \).
- Thus:
\[ \int x \cos x \, dx = x \sin x - \int \sin x \, dx \]
- The integral of \( \sin x \) is:
\[ \int \sin x \, dx = -\cos x \]
- Therefore:
\[ \int x \cos x \, dx = x \sin x + \cos x \]

4. **Substitute back**:
- Now substitute back into the first integral:
\[ \int x^2 \sin x \, dx = -x^2 \cos x + 2(x \sin x + \cos x) \]
- Simplifying gives:
\[ = -x^2 \cos x + 2x \sin x + 2 \cos x + C \]

### Final Result for (i):
\[ \int x^2 \sin x \, dx = -x^2 \cos x + 2x \sin x + 2 \cos x + C \]

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### (ii) Integral of \( \int x^2 \sin 2x \, dx \)

1. **Choose \( u \) and \( dv \)**:
- Let \( u = x^2 \) and \( dv = \sin 2x \, dx \).
- Then, differentiate \( u \) to find \( du \):
\[ du = 2x \, dx \]
- Integrate \( dv \) to find \( v \):
\[ v = -\frac{1}{2} \cos 2x \]

2. **Apply Integration by Parts**:
- Using the formula \( \int u \, dv = uv - \int v \, du \):
\[ \int x^2 \sin 2x \, dx = -\frac{1}{2} x^2 \cos 2x - \int -\frac{1}{2} \cos 2x (2x) \, dx \]
- Simplifying gives:
\[ = -\frac{1}{2} x^2 \cos 2x + \int x \cos 2x \, dx \]

3. **Integrate \( \int x \cos 2x \, dx \)**:
- Again, apply integration by parts:
- Let \( u = x \) and \( dv = \cos 2x \, dx \).
- Then, \( du = dx \) and \( v = \frac{1}{2} \sin 2x \).
- Thus:
\[ \int x \cos 2x \, dx = \frac{1}{2} x \sin 2x - \int \frac{1}{2} \sin 2x \, dx \]
- The integral of \( \sin 2x \) is:
\[ \int \sin 2x \, dx = -\frac{1}{2} \cos 2x \]
- Therefore:
\[ \int x \cos 2x \, dx = \frac{1}{2} x \sin 2x + \frac{1}{4} \cos 2x \]

4. **Substitute back**:
- Now substitute back into the second integral:
\[ \int x^2 \sin 2x \, dx = -\frac{1}{2} x^2 \cos 2x + \left( \frac{1}{2} x \sin 2x + \frac{1}{4} \cos 2x \right) \]
- Simplifying gives:
\[ = -\frac{1}{2} x^2 \cos 2x + \frac{1}{2} x \sin 2x + \frac{1}{4} \cos 2x + C \]

### Final Result for (ii):
\[ \int x^2 \sin 2x \, dx = -\frac{1}{2} x^2 \cos 2x + \frac{1}{2} x \sin 2x + \frac{1}{4} \cos 2x + C \]

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