Q.Integrate the following function: $\sin x \sin 2x \sin 3x$

Reduce the triple product to a sum using product-to-sum twice.

First $\sin x\sin 2x=\frac12[\cos x-\cos 3x]$, so

Now $\cos x\sin 3x=\frac12[\sin 4x+\sin 2x]$ and $\cos 3x\sin 3x=\frac12\sin 6x$, giving

Integrate term by term: …

Plan

A product of three sines can't be integrated as is, so peel it into a sum by applying product-to-sum identities twice.

Step 1: pair two factors

(using $\cos(-x)=\cos x$). So

Step 2: expand each product

$\cos x\sin 3x=\frac12[\sin(3x+x)-\sin(x-3x)]=\frac12[\sin 4x+\sin 2x]$ (since $\sin(-2x)=-\sin 2x$), and $\cos 3x\sin 3x=\frac12\sin 6x$. Hence …

Method: Product-to-sum applied twice for a triple sine product

A product of three sines integrates only after being turned into a sum. Reduce two factors first, distribute the third, reduce again, then integrate each single-angle term.

Steps

Step 1: Combine two of the sines.

For example $\sin x\sin 2x=\tfrac12[\cos x-\cos 3x]$.

Step 2: Multiply by the third factor and reduce once more. …

Common Mistakes

Mistake 1: Using the wrong sign or order in the $\sin\sin$ identity.

Why it's wrong: the correct form is $\sin A\sin B=\tfrac12[\cos(A-B)-\cos(A+B)]$ — $\cos(A-B)$ first, then a minus. Swapping them flips signs on the final cosines. Correct approach: apply the identity exactly as stated at each of the two reduction steps.

Mistake 2: Attempting to integrate the triple product directly. …