Evaluate the integral: I = integral sin (2x- pi/6) cos
Evaluate the integral: $$ I = \int \sin\!\left(2x-\frac{\pi}{6}\right)\cos\!\left(3x+\frac{\pi}{4}\right) dx. $$
Explanation of Method
Even with phase shifts, the product-to-sum identity $\sin P\cos Q=\tfrac12[\sin(P+Q)+\sin(P-Q)]$ applies directly — the phase constants simply carry through the algebra of combining $P\pm Q$.
- Apply the product-to-sum identity.
- Simplify the resulting phase-shifted 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 $P=2x-\tfrac{\pi}{6}$, $Q=3x+\tfrac{\pi}{4}$:
$$ \sin\!\left(2x-\frac{\pi}{6}\right)\cos\!\left(3x+\frac{\pi}{4}\right) = \frac12\left[\sin(P+Q) + \sin(P-Q)\right]. $$
Step 2 · Simplify the arguments
$$ P+Q = 5x + \left(\frac{\pi}{4}-\frac{\pi}{6}\right) = 5x+\frac{\pi}{12}, \qquad P-Q = -x - \left(\frac{\pi}{6}+\frac{\pi}{4}\right) = -x-\frac{5\pi}{12}. $$
Using $\sin(-\theta)=-\sin\theta$:
$$ \sin\!\left(2x-\frac{\pi}{6}\right)\cos\!\left(3x+\frac{\pi}{4}\right) = \frac12\left[\sin\!\left(5x+\frac{\pi}{12}\right) - \sin\!\left(x+\frac{5\pi}{12}\right)\right]. $$
Step 3 · Integrate each term
Using $\int \sin(ax+b)\,dx = -\tfrac1a\cos(ax+b)$:
$$ \int \sin\!\left(5x+\frac{\pi}{12}\right) dx = -\frac15\cos\!\left(5x+\frac{\pi}{12}\right), \qquad \int \sin\!\left(x+\frac{5\pi}{12}\right) dx = -\cos\!\left(x+\frac{5\pi}{12}\right). $$
Step 4 · Combine the results
$$ I = \frac12\left[-\frac15\cos\!\left(5x+\frac{\pi}{12}\right) + \cos\!\left(x+\frac{5\pi}{12}\right)\right] + C = -\frac{1}{10}\cos\!\left(5x+\frac{\pi}{12}\right) + \frac12\cos\!\left(x+\frac{5\pi}{12}\right) + C. $$
Summary Table
| Term | Antiderivative |
|---|---|
| $\tfrac12\sin(5x+\pi/12)$ | $-\tfrac1{10}\cos(5x+\pi/12)$ |
| $-\tfrac12\sin(x+5\pi/12)$ | $\tfrac12\cos(x+5\pi/12)$ |
Additional Notes
- The integrand is defined for all real $x$.
- Carrying the phase constants ($\pi/6$, $\pi/4$) through the identity is purely bookkeeping — the structure of the calculation is identical to Problem 126's fractional-frequency case, just with fixed offsets added to each argument.
- The result can be checked by differentiating the final expression and confirming it returns the original product $\sin(2x-\pi/6)\cos(3x+\pi/4)$.
Result
The solution to the integral is: