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