Evaluate the integral: I = integral sin x sin2x sin3x dx

Evaluate the integral: $$ I = \int \sin x\,\sin2x\,\sin3x\,dx. $$

Step 1 · Combine the first two factors

Using $\sin P\sin Q=\tfrac12[\cos(P-Q)-\cos(P+Q)]$ with $P=x$, $Q=2x$:

$$ \sin x\sin2x = \frac12\left[\cos(-x)-\cos3x\right] = \frac12\left[\cos x-\cos3x\right]. $$

So the integrand becomes:

$$ \sin x\sin2x\sin3x = \frac12\left(\cos x-\cos3x\right)\sin3x = \frac12\left[\cos x\sin3x - \cos3x\sin3x\right]. $$

Step 2 · Expand each product

For the first piece, using $\cos P\sin Q=\tfrac12[\sin(P+Q)-\sin(P-Q)]$ with $P=x$, $Q=3x$:

$$ \cos x\sin3x = \frac12\left[\sin4x-\sin(-2x)\right] = \frac12\left[\sin4x+\sin2x\right]. $$

For the second piece, using the double-angle identity $\sin\alpha\cos\alpha=\tfrac12\sin2\alpha$ with $\alpha=3x$:

$$ \cos3x\sin3x = \frac12\sin6x. $$

Substituting both back:

$$ \sin x\sin2x\sin3x = \frac12\left[\frac12(\sin4x+\sin2x) - \frac12\sin6x\right] = \frac14\left[\sin4x+\sin2x-\sin6x\right]. $$

Step 3 · Integrate term by term

$$ I = \frac14\int \left(\sin4x+\sin2x-\sin6x\right) dx = \frac14\left[-\frac14\cos4x - \frac12\cos2x + \frac16\cos6x\right] + C. $$

Step 4 · Simplify

$$ I = -\frac{1}{16}\cos4x - \frac18\cos2x + \frac{1}{24}\cos6x + C. $$

Summary Table

Additional Notes

• The integrand is defined for all real $x$.
• A common pitfall when chaining product-to-sum identities like this is mixing up the sign convention for $\cos P\sin Q$ versus $\sin P\cos Q$ at each stage — here the $\cos x\sin3x$ step must use the minus-sign form, exactly as in Problem 123.
• The result can be checked by differentiating $-\tfrac{1}{16}\cos4x-\tfrac18\cos2x+\tfrac{1}{24}\cos6x$ and confirming it returns $\sin x\sin2x\sin3x$.