Evaluate the integral: I = integral sin3x sin5x dx

Evaluate the integral: $$ I = \int \sin3x\,\sin5x\,dx. $$

Step 1 · Apply the product-to-sum identity

Using $\sin P\sin Q = \tfrac12[\cos(P-Q)-\cos(P+Q)]$ with $P=5x$, $Q=3x$:

$$ \sin3x\sin5x = \frac12\left[\cos(5x-3x) - \cos(5x+3x)\right] = \frac12\left[\cos2x-\cos8x\right]. $$

Step 2 · Rewrite the integral

$$ I = \frac12\int \cos2x\,dx - \frac12\int \cos8x\,dx. $$

Step 3 · Integrate each term

$$ \int \cos2x\,dx = \frac12\sin2x, \qquad \int \cos8x\,dx = \frac18\sin8x. $$

Step 4 · Combine the results

$$ I = \frac12\left(\frac12\sin2x\right) - \frac12\left(\frac18\sin8x\right) + C = \frac14\sin2x - \frac{1}{16}\sin8x + C. $$

Summary Table

Additional Notes

• The integrand is defined for all real $x$.
• Since $\sin P\sin Q=\sin Q\sin P$, the identity can be applied with either factor as $P$; the sign of the $\cos(P-Q)$ term is unaffected because cosine is even.
• The result can be checked by differentiating $\tfrac14\sin2x-\tfrac{1}{16}\sin8x$ and confirming it returns $\sin3x\sin5x$.