3. Evaluate the following integrals by method of partial fractions $$ \begin{array}{ll} ext { (a) } \int \cos 6 x \cos 4 x d x & ext { (b) } \int \sin 5 x \sin x d x\end{array} $$

Question

3. Evaluate the following integrals by method of partial fractions $$ \begin{array}{ll}	ext { (a) } \int \cos 6 x \cos 4 x d x & 	ext { (b) } \int \sin 5 x \sin x d x\end{array} $$

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To evaluate the integrals using trigonometric identities (often referred to as the method of partial fractions in this context), we'll employ product-to-sum formulas. These formulas simplify the product of trigonometric functions into a sum, making the integration straightforward.

### Problem (a)
$$ \int \cos 6x \cos 4x \, dx $$

**Solution:**

Use the product-to-sum identity:
$$
\cos A \cos B = \frac{1}{2} [\cos(A + B) + \cos(A - B)]
$$

Applying this to the integral:
$$
\cos 6x \cos 4x = \frac{1}{2} [\cos(10x) + \cos(2x)]
$$

So, the integral becomes:
$$
\int \cos 6x \cos 4x \, dx = \frac{1}{2} \int \cos 10x \, dx + \frac{1}{2} \int \cos 2x \, dx
$$

Integrate term by term:
$$
\frac{1}{2} \int \cos 10x \, dx = \frac{1}{2} \cdot \frac{\sin 10x}{10} = \frac{\sin 10x}{20}
$$
$$
\frac{1}{2} \int \cos 2x \, dx = \frac{1}{2} \cdot \frac{\sin 2x}{2} = \frac{\sin 2x}{4}
$$

Combine the results:
$$
\int \cos 6x \cos 4x \, dx = \frac{\sin 10x}{20} + \frac{\sin 2x}{4} + C
$$
where $$ C $$ is the constant of integration.

### Problem (b)
$$ \int \sin 5x \sin x \, dx $$

**Solution:**

Use the product-to-sum identity:
$$
\sin A \sin B = \frac{1}{2} [\cos(A - B) - \cos(A + B)]
$$

Applying this to the integral:
$$
\sin 5x \sin x = \frac{1}{2} [\cos(4x) - \cos(6x)]
$$

So, the integral becomes:
$$
\int \sin 5x \sin x \, dx = \frac{1}{2} \int \cos 4x \, dx - \frac{1}{2} \int \cos 6x \, dx
$$

Integrate term by term:
$$
\frac{1}{2} \int \cos 4x \, dx = \frac{1}{2} \cdot \frac{\sin 4x}{4} = \frac{\sin 4x}{8}
$$
$$
\frac{1}{2} \int \cos 6x \, dx = \frac{1}{2} \cdot \frac{\sin 6x}{6} = \frac{\sin 6x}{12}
$$

Combine the results:
$$
\int \sin 5x \sin x \, dx = \frac{\sin 4x}{8} - \frac{\sin 6x}{12} + C
$$
where $$ C $$ is the constant of integration.

---

**Final Answers:**

**a.**
$$
\int \cos 6x \cos 4x \, dx = \frac{\sin 10x}{20} + \frac{\sin 2x}{4} + C
$$

**b.**
$$
\int \sin 5x \sin x \, dx = \frac{\sin 4x}{8} - \frac{\sin 6x}{12} + C
$$
**a.** $$ \int \cos 6x \cos 4x \, dx = \frac{\sin 10x}{20} + \frac{\sin 2x}{4} + C $$ **b.** $$ \int \sin 5x \sin x \, dx = \frac{\sin 4x}{8} - \frac{\sin 6x}{12} + C $$

Evaluate the following integrals by method of partial fractions

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