`(i) int sin 2x. cos5x dx " "(ii) int(sin 4x)/(sin x) dx`

Answer

Let's solve the given integrals step by step.

### Part (i): \( \int \sin(2x) \cos(5x) \, dx \)

1. **Use the product-to-sum identities**:
We can use the identity:
\[ \sin A \cos B = \frac{1}{2} [\sin(A + B) + \sin(A - B)] \]
Here, \( A = 2x \) and \( B = 5x \). Thus,
\[ \sin(2x) \cos(5x) = \frac{1}{2} [\sin(2x + 5x) + \sin(2x - 5x)] = \frac{1}{2} [\sin(7x) + \sin(-3x)] \]
Since \( \sin(-\theta) = -\sin(\theta) \), we have:
\[ \sin(-3x) = -\sin(3x) \]
Therefore,
\[ \sin(2x) \cos(5x) = \frac{1}{2} [\sin(7x) - \sin(3x)] \]

2. **Integrate**:
Now, we can write the integral as:
\[ \int \sin(2x) \cos(5x) \, dx = \frac{1}{2} \int [\sin(7x) - \sin(3x)] \, dx \]
This can be split into two separate integrals:
\[ = \frac{1}{2} \left( \int \sin(7x) \, dx - \int \sin(3x) \, dx \right) \]

3. **Calculate the integrals**:
The integrals are:
\[ \int \sin(7x) \, dx = -\frac{1}{7} \cos(7x) + C_1 \]
\[ \int \sin(3x) \, dx = -\frac{1}{3} \cos(3x) + C_2 \]
Therefore,
\[ = \frac{1}{2} \left( -\frac{1}{7} \cos(7x) + \frac{1}{3} \cos(3x) \right) + C \]
Simplifying gives:
\[ = -\frac{1}{14} \cos(7x) + \frac{1}{6} \cos(3x) + C \]

### Final Answer for Part (i):
\[ \int \sin(2x) \cos(5x) \, dx = -\frac{1}{14} \cos(7x) + \frac{1}{6} \cos(3x) + C \]

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### Part (ii): \( \int \frac{\sin(4x)}{\sin(x)} \, dx \)

1. **Rewrite \( \sin(4x) \)**:
We can use the identity:
\[ \sin(4x) = 2 \sin(2x) \cos(2x) \]
Thus,
\[ \int \frac{\sin(4x)}{\sin(x)} \, dx = \int \frac{2 \sin(2x) \cos(2x)}{\sin(x)} \, dx \]

2. **Use the double angle formula for \( \sin(2x) \)**:
We know that:
\[ \sin(2x) = 2 \sin(x) \cos(x) \]
So,
\[ \int \frac{2(2 \sin(x) \cos(x)) \cos(2x)}{\sin(x)} \, dx = \int 4 \cos(x) \cos(2x) \, dx \]

3. **Use the product-to-sum identities again**:
Using the identity:
\[ \cos A \cos B = \frac{1}{2} [\cos(A + B) + \cos(A - B)] \]
Here, \( A = x \) and \( B = 2x \):
\[ \cos(x) \cos(2x) = \frac{1}{2} [\cos(3x) + \cos(x)] \]
Therefore,
\[ 4 \cos(x) \cos(2x) = 2 [\cos(3x) + \cos(x)] \]

4. **Integrate**:
Now we can write the integral as:
\[ \int 2 [\cos(3x) + \cos(x)] \, dx = 2 \left( \int \cos(3x) \, dx + \int \cos(x) \, dx \right) \]

5. **Calculate the integrals**:
\[ \int \cos(3x) \, dx = \frac{1}{3} \sin(3x) + C_1 \]
\[ \int \cos(x) \, dx = \sin(x) + C_2 \]
Thus,
\[ = 2 \left( \frac{1}{3} \sin(3x) + \sin(x) \right) + C \]
Simplifying gives:
\[ = \frac{2}{3} \sin(3x) + 2 \sin(x) + C \]

### Final Answer for Part (ii):
\[ \int \frac{\sin(4x)}{\sin(x)} \, dx = \frac{2}{3} \sin(3x) + 2 \sin(x) + C \]

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