`(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): \( \int \sin(2x) \cos(5x) \, dx \)

**Step 1: Use the product-to-sum identities.**
We can rewrite the product of sine and cosine using the identity:
\[ \sin A \cos B = \frac{1}{2} [\sin(A + B) + \sin(A - B)] \]
Here, \( A = 2x \) and \( B = 5x \).

So,
\[ \sin(2x) \cos(5x) = \frac{1}{2} [\sin(2x + 5x) + \sin(2x - 5x)] = \frac{1}{2} [\sin(7x) + \sin(-3x)] \]

**Step 2: Simplify using the property of sine.**
Since \( \sin(-\theta) = -\sin(\theta) \), we have:
\[ \sin(-3x) = -\sin(3x) \]
Thus,
\[ \sin(2x) \cos(5x) = \frac{1}{2} [\sin(7x) - \sin(3x)] \]

**Step 3: Substitute back into the integral.**
Now we can rewrite the integral:
\[ \int \sin(2x) \cos(5x) \, dx = \frac{1}{2} \int [\sin(7x) - \sin(3x)] \, dx \]

**Step 4: Integrate each term.**
Now we integrate:
\[ \int \sin(7x) \, dx = -\frac{1}{7} \cos(7x) + C_1 \]
\[ \int \sin(3x) \, dx = -\frac{1}{3} \cos(3x) + C_2 \]

Putting it all together:
\[ \int \sin(2x) \cos(5x) \, dx = \frac{1}{2} \left(-\frac{1}{7} \cos(7x) + \frac{1}{3} \cos(3x)\right) + C \]

**Step 5: Simplify the expression.**
This gives us:
\[ = -\frac{1}{14} \cos(7x) + \frac{1}{6} \cos(3x) + C \]

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

**Step 1: Rewrite \(\sin(4x)\) using the double angle formula.**
We know that:
\[ \sin(4x) = 2 \sin(2x) \cos(2x) \]
Thus, we can rewrite the integral as:
\[ \int \frac{2 \sin(2x) \cos(2x)}{\sin(x)} \, dx \]

**Step 2: Use the identity for \(\sin(2x)\).**
Recall that:
\[ \sin(2x) = 2 \sin(x) \cos(x) \]
So, substituting this into our integral gives:
\[ \int \frac{2(2 \sin(x) \cos(x)) \cos(2x)}{\sin(x)} \, dx = \int 4 \cos(x) \cos(2x) \, dx \]

**Step 3: Use the product-to-sum identities again.**
Using the identity:
\[ \cos A \cos B = \frac{1}{2} [\cos(A + B) + \cos(A - B)] \]
we have:
\[ \cos(x) \cos(2x) = \frac{1}{2} [\cos(3x) + \cos(x)] \]

**Step 4: Substitute back into the integral.**
Now we can rewrite the integral:
\[ \int 4 \cos(x) \cos(2x) \, dx = 4 \cdot \frac{1}{2} \int [\cos(3x) + \cos(x)] \, dx = 2 \int [\cos(3x) + \cos(x)] \, dx \]

**Step 5: Integrate each term.**
Now we integrate:
\[ \int \cos(3x) \, dx = \frac{1}{3} \sin(3x) + C_1 \]
\[ \int \cos(x) \, dx = \sin(x) + C_2 \]

Putting it all together:
\[ 2 \int [\cos(3x) + \cos(x)] \, dx = 2 \left(\frac{1}{3} \sin(3x) + \sin(x)\right) + C \]

**Step 6: Simplify the expression.**
This gives us:
\[ = \frac{2}{3} \sin(3x) + 2 \sin(x) + C \]

### 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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