Evaluate the integral: I = integral sin x sin(x+ theta) dx
Evaluate the integral: $$ I = \int \sin x\,\sin(x+\theta)\,dx. $$
Explanation of Method
Treating $\theta$ as a constant, the product-to-sum identity $\sin P\sin Q=\tfrac12[\cos(P-Q)-\cos(P+Q)]$ turns the product into a constant term plus a single cosine term in $x$.
- Apply the product-to-sum identity with $P=x$, $Q=x+\theta$.
- Simplify using $\cos(-\theta)=\cos\theta$.
- Integrate each term and combine.
Step 1 · Apply the product-to-sum identity
Using $\sin P\sin Q = \tfrac12[\cos(P-Q)-\cos(P+Q)]$ with $P=x$, $Q=x+\theta$:
$$ \sin x\sin(x+\theta) = \frac12\left[\cos(x-(x+\theta)) - \cos(x+(x+\theta))\right] = \frac12\left[\cos(-\theta) - \cos(2x+\theta)\right]. $$
Step 2 · Simplify
Since $\cos(-\theta)=\cos\theta$:
$$ \sin x\sin(x+\theta) = \frac12\left[\cos\theta - \cos(2x+\theta)\right]. $$
Step 3 · Integrate
Substituting into the integral, and treating $\theta$ as constant:
$$ I = \frac12\cos\theta\int dx - \frac12\int \cos(2x+\theta)\,dx = \frac{x}{2}\cos\theta - \frac12\left(\frac12\sin(2x+\theta)\right) + C. $$
Step 4 · Simplify
$$ I = \frac{x}{2}\cos\theta - \frac14\sin(2x+\theta) + C. $$
Summary Table
| Term | Antiderivative |
|---|---|
| $\tfrac12\cos\theta$ (constant) | $\tfrac{x}{2}\cos\theta$ |
| $-\tfrac12\cos(2x+\theta)$ | $-\tfrac14\sin(2x+\theta)$ |
Additional Notes
- The integrand is defined for all real $x$ and any constant $\theta$.
- The key simplification is recognizing that $\cos(x-(x+\theta))=\cos(-\theta)$ is a constant with respect to $x$, so it integrates to a linear term $\tfrac{x}{2}\cos\theta$ rather than another trig function.
- The result can be checked by differentiating $\tfrac{x}{2}\cos\theta-\tfrac14\sin(2x+\theta)$ with respect to $x$ and confirming it returns $\sin x\sin(x+\theta)$.
Result
The solution to the integral is: