Q.Integrate the following function: $\cos 2x \cos 4x \cos 6x$

Concept: Product-to-Sum Identity — repeatedly convert products of cosines into sums to make integration straightforward.

First, pair $\cos 2x \cos 4x$ using the identity

$\cos A \cos B = \frac{1}{2}[\cos(A+B) + \cos(A-B)]$:

Now multiply by $\cos 6x$:

Apply the identity again to $\cos 2x \cos 6x = \frac{1}{2}[\cos 8x + \cos 4x]$, and use $\cos^2 6x = \frac{1 + \cos 12x}{2}$:

Integrate term by term:

When you see a product of three cosines, your first instinct might be to try a substitution or a trigonometric identity like $\cos 2x = 2\cos^2 x - 1$. That would lead to a messy polynomial in cosines — doable, but unnecessarily long. The cleanest path is the product-to-sum identity, because it converts multiplication into addition, and addition is trivial to integrate.

The identity $\cos A \cos B = \frac{1}{2}[\cos(A+B) + \cos(A-B)]$ is your workhorse here. You apply it pairwise, one pair at a time. The order matters only for convenience — we’ll start with $\cos 2x$ and $\cos 4x$.

1. First product-to-sum step Take the first two factors:

Since cosine is even, $\cos(-2x) = \cos 2x$. So:

2. Multiply by the third factor Now multiply this result by $\cos 6x$:

3. Handle $\cos^2 6x$ Use the double-angle identity: $\cos^2 \theta = \frac{1 + \cos 2\theta}{2}$. Here $\theta = 6x$, so:

4. Handle $\cos 2x \cos 6x$ Apply product-to-sum again:

5. Combine everything Substitute back:

Factor the $\frac{1}{2}$ outside:

6. Integrate term-by-term Now integrate:

Each term is straightforward:

• $\int 1 \, dx = x$
• $\int \cos 12x \, dx = \frac{\sin 12x}{12}$
• $\int \cos 8x \, dx = \frac{\sin 8x}{8}$
• $\int \cos 4x \, dx = \frac{\sin 4x}{4}$

So:

Method: Product-to-sum applied repeatedly for a triple cosine product

For a product of three cosines (or sines), collapse two factors at a time with product-to-sum identities until only single-angle terms remain, then integrate.

Steps

Step 1: Combine any two of the three factors.

Pick a convenient pair (e.g. $\cos 2x\cos 6x$) to turn the triple product into a cosine times a sum.

Step 2: Distribute the remaining factor and reduce again.

Multiplying the leftover cosine through gives more products (including a $\cos^2$ term). Apply product-to-sum a second time, and use $\cos^2\theta=\dfrac{1+\cos 2\theta}{2}$ for any squared cosine that appears.

Step 3: Integrate the resulting single-angle terms.

You end with a constant plus several $\cos(kx)$ terms. Integrate using $\displaystyle\int\cos(kx)\,dx=\frac1k\sin(kx)$, and the constant gives an $x$ term.

The key discipline is bookkeeping: reduce one pair fully before touching the third factor, so no cross term is lost.

Common Mistakes

Mistake 1: Dropping the $\cos^2$ term that appears mid-way.

Why it's wrong: after the first product-to-sum step a $\cos^2$ (or a repeated angle) shows up and must itself be reduced with $\cos^2\theta=\tfrac{1+\cos 2\theta}{2}$; ignoring it loses the constant term that produces the $x$ in the answer. Correct approach: reduce every squared/repeated factor before integrating.

Mistake 2: Forgetting the $\dfrac1k$ factors when integrating each $\cos(kx)$.

Why it's wrong: each term integrates as $\dfrac{\sin(kx)}{k}$, so $\cos 12x$, $\cos 8x$, $\cos 4x$ carry different divisors $12,8,4$. Using a single divisor corrupts every coefficient. Correct approach: divide each cosine by its own angular coefficient.