Q.Integrate the following function: $\sin 4x \sin 8x$

The key idea is to use the Product-to-Sum identity to rewrite the product of sines as a sum of cosines, which is straightforward to integrate.

Step 1: Recall the identity:

Step 2: Apply it with $A = 4x$ and $B = 8x$:

Step 3: Since $\cos(-4x) = \cos 4x$, the expression simplifies to:

$\sin 4x \sin 8x = \frac{1}{2} \left( \cos 4x - \cos 12x \right)$ …

When you see a product of two sine functions (or sine and cosine), the direct approach — trying to guess a reverse chain rule — fails because the angles are different. The trick is to rewrite the product as a sum or difference of cosines. This is one of the Product-to-Sum identities, and it exists precisely to turn multiplication (hard to integrate) into addition (easy to integrate).

The identity we need is:

Why does this work? Because the derivative of $\sin$ is $\cos$, and the derivative of $\cos$ is $-\sin$ — so integrating a cosine is straightforward. By converting the product into a combination of cosines, each term becomes a basic integral.

Let’s apply it step by step.

1. Identify $A$ and $B$. Here, $A = 4x$ and $B = 8x$. Plug into the identity:

2. Simplify the angles inside the cosines. $4x - 8x = -4x$, and $\cos(-4x) = \cos 4x$ because cosine is an even function. $4x + 8x = 12x$. So:

3. Set up the integral.

4. Integrate each cosine term separately.

   Recall: $\int \cos(kx) \, dx = \frac{1}{k} \sin(kx) + C$.

   So:

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

   Therefore: …

Method: Product-to-sum for $\sin(ax)\sin(bx)$

Convert a product of two sines into a difference of cosines, then integrate term by term.

Steps

Step 1: Apply the sine-sine identity.

Match $A$ and $B$ to the two angles.

Step 2: Simplify negative angles.

Cosine is even, so $\cos(A-B)=\cos(B-A)$ — a negative angle inside the cosine can be flipped without a sign change.

Step 3: Integrate each cosine. …

Common Mistakes

Mistake 1: Dropping the $\tfrac12$ factor from the product-to-sum identity.

Why it's wrong: $\sin 4x\sin 8x=\tfrac12[\cos 4x-\cos 12x]$; omitting the $\tfrac12$ doubles every term of the answer. Correct approach: carry the $\tfrac12$ through, giving $\tfrac18\sin 4x-\tfrac{1}{24}\sin 12x+C$.

Mistake 2: Writing the identity with a plus instead of a minus between the cosines. …