Evaluate the integral: I = integral cos x/2 cos x/3 dx

Evaluate the integral: $$ I = \int \cos\frac{x}{2}\,\cos\frac{x}{3}\,dx. $$

Step 1 · Apply the product-to-sum identity

Using $\cos P\cos Q = \tfrac12[\cos(P-Q)+\cos(P+Q)]$ with $P=\tfrac{x}{2}$, $Q=\tfrac{x}{3}$:

$$ \cos\frac{x}{2}\cos\frac{x}{3} = \frac12\left[\cos\left(\frac{x}{2}-\frac{x}{3}\right) + \cos\left(\frac{x}{2}+\frac{x}{3}\right)\right]. $$

Step 2 · Simplify the arguments

$$ \frac{x}{2}-\frac{x}{3} = \frac{3x}{6}-\frac{2x}{6} = \frac{x}{6}, \qquad \frac{x}{2}+\frac{x}{3} = \frac{3x}{6}+\frac{2x}{6} = \frac{5x}{6}. $$

So:

$$ I = \frac12\int \cos\frac{x}{6}\,dx + \frac12\int \cos\frac{5x}{6}\,dx. $$

Step 3 · Integrate each term

Using $\int \cos kx\,dx = \tfrac1k\sin kx$:

$$ \int \cos\frac{x}{6}\,dx = 6\sin\frac{x}{6}, \qquad \int \cos\frac{5x}{6}\,dx = \frac65\sin\frac{5x}{6}. $$

Step 4 · Combine the results

$$ I = \frac12\left(6\sin\frac{x}{6}+\frac65\sin\frac{5x}{6}\right) + C = 3\sin\frac{x}{6} + \frac35\sin\frac{5x}{6} + C. $$

Summary Table

Additional Notes

• The integrand is defined for all real $x$.
• Small denominators in the frequencies, like the $\tfrac16$ here, produce comparatively large coefficients in the antiderivative (like the $6$ in front of $\sin\tfrac{x}{6}$) — a useful sanity check when reviewing the result.
• The result can be checked by differentiating $3\sin\tfrac{x}{6}+\tfrac35\sin\tfrac{5x}{6}$ and confirming it returns $\cos\tfrac{x}{2}\cos\tfrac{x}{3}$.