Evaluate the integral: I = integral sin x/2 cos x/12 dx
Evaluate the integral: $$ I = \int \sin\frac{x}{2}\,\cos\frac{x}{12}\,dx. $$
Explanation of Method
The product-to-sum identity $\sin P\cos Q=\tfrac12[\sin(P+Q)+\sin(P-Q)]$ applies just as well to fractional-frequency arguments, converting the product into a sum of two single sine terms.
- Apply the product-to-sum identity.
- Simplify the resulting arguments.
- Integrate each term and combine.
Step 1 · Apply the product-to-sum identity
Using $\sin P\cos Q = \tfrac12[\sin(P+Q)+\sin(P-Q)]$ with $P=\tfrac{x}{2}$, $Q=\tfrac{x}{12}$:
$$ \sin\frac{x}{2}\cos\frac{x}{12} = \frac12\left[\sin\left(\frac{x}{2}+\frac{x}{12}\right) + \sin\left(\frac{x}{2}-\frac{x}{12}\right)\right]. $$
Step 2 · Simplify the arguments
$$ \frac{x}{2}+\frac{x}{12} = \frac{6x}{12}+\frac{x}{12} = \frac{7x}{12}, \qquad \frac{x}{2}-\frac{x}{12} = \frac{6x}{12}-\frac{x}{12} = \frac{5x}{12}. $$
So:
$$ I = \frac12\int \sin\frac{7x}{12}\,dx + \frac12\int \sin\frac{5x}{12}\,dx. $$
Step 3 · Integrate each term
Using $\int \sin kx\,dx = -\tfrac1k\cos kx$:
$$ \int \sin\frac{7x}{12}\,dx = -\frac{12}{7}\cos\frac{7x}{12}, \qquad \int \sin\frac{5x}{12}\,dx = -\frac{12}{5}\cos\frac{5x}{12}. $$
Step 4 · Combine the results
$$ I = \frac12\left(-\frac{12}{7}\cos\frac{7x}{12}-\frac{12}{5}\cos\frac{5x}{12}\right) + C = -\frac67\cos\frac{7x}{12} - \frac65\cos\frac{5x}{12} + C. $$
Summary Table
| Term | Antiderivative |
|---|---|
| $\tfrac12\sin\tfrac{7x}{12}$ | $-\tfrac67\cos\tfrac{7x}{12}$ |
| $\tfrac12\sin\tfrac{5x}{12}$ | $-\tfrac65\cos\tfrac{5x}{12}$ |
Additional Notes
- The integrand is defined for all real $x$.
- The product-to-sum identities work identically regardless of whether the frequencies are integers or fractions — only the arithmetic of combining $P\pm Q$ changes.
- The result can be checked by differentiating $-\tfrac67\cos\tfrac{7x}{12}-\tfrac65\cos\tfrac{5x}{12}$ and confirming it returns $\sin\tfrac{x}{2}\cos\tfrac{x}{12}$.
Result
The solution to the integral is: