The given function is \(\displaystyle \sin^{4} x \) and we need to compute the integral of this function with respect to \( x \), i.e., compute \(\displaystyle \int \sin^{4} x \, dx \).
Solution:
\( \begin{aligned} &\int \sin^{4} x \, dx \\ &= \int \left(\frac{1 – \cos 2x}{2}\right)^{2} \, dx, \quad \because \sin^{2} x = \frac{1 – \cos 2x}{2} \\ &= \int \frac{1 – 2\cos 2x + \cos^{2} 2x}{4} \, dx \\ &= \int \frac{1 – 2\cos 2x + \frac{1 + \cos 4x}{2}}{4} \, dx, \quad \because \cos^{2} 2x = \frac{1 + \cos 4x}{2} \\ &= \int \frac{2 – 4\cos 2x + 1 + \cos 4x}{8} \, dx \\ &= \int \frac{3 – 4\cos 2x + \cos 4x}{8} \, dx \\ &= \int \frac{3}{8} \, dx \, – \int \frac{4\cos 2x}{8} \, dx + \int \frac{\cos 4x}{8} \, dx \\ &= \frac{3}{8}x \, – \frac{1}{2} \cdot \frac{\sin 2x}{2} + \frac{1}{8} \cdot \frac{\sin 4x}{4} + C, \quad \text{C is a constant} \\ &= \frac{3}{8}x \, – \frac{\sin 2x}{4} + \frac{\sin 4x}{32} + C \end{aligned} \)Thus, the integral of the given function is
\( \boxed{\int \sin^{4} x \, dx = \frac{3}{8}x \,- \frac{\sin 2x}{4} + \frac{\sin 4x}{32} + C} \\ \)Please let me know in the comments if you find any errors in this solution.