The given function is \(\displaystyle \sin 3x \cdot \sin 5x \) and we need to compute the integral of this function with respect to \( x \), i.e., compute \(\displaystyle \int \sin 3x \cdot \sin 5x \, dx \).
Solution:
Using the product-to-sum identity:
\( \sin A \sin B = \frac{1}{2} [\cos(A – B) – \cos(A + B)] \)
\( \begin{aligned} &\int \sin 3x \cdot \sin 5x \, dx \\ &= \int \frac{1}{2} [\cos(3x – 5x) \,- \cos(3x + 5x)] \, dx \\ &= \frac{1}{2} [\cos(-2x) \,- \cos(8x)] \, dx \\ &= \int \frac{1}{2} [\cos(2x) \,- \cos(8x)] \, dx \quad \because \cos(-\theta) = \cos \theta \\ &= \frac{1}{2} \int [\cos(2x) \,- \cos(8x)] \, dx \\ &= \frac{1}{2} \int \cos(2x) \, dx \,- \frac{1}{2} \int \cos(8x) \, dx \\ &= \frac{1}{2} \left( \frac{1}{2} \sin(2x) \,- \frac{1}{8} \sin(8x) \right) + C, \quad \text{C is a constant} \\ &= \frac{1}{4} \sin(2x) \,- \frac{1}{16} \sin(8x) + C \\ \end{aligned} \)Thus, the integral of the given function is
\( \boxed{\int \sin 3x \cdot \sin 5x \, dx = \frac{1}{4} \sin(2x) \,- \frac{1}{16} \sin(8x) + C} \)Please let me know in the comments if you find any errors in this solution.