​$$\int_{0}^{\frac{\pi}{4}} \sin^3 \theta \, d\theta = \, ?$$​​

Correct option is A

​$$\textbf{Use the identity } \sin^3 \theta = \sin \theta(1 - \cos^2 \theta) \\\int \sin^3 \theta \, d\theta = \int \sin \theta(1 - \cos^2 \theta) \, d\theta \\\textbf{Let } u = \cos \theta, \, du = - \sin \theta \, d\theta \\\int \sin \theta(1 - \cos^2 \theta) \, d\theta = \int (1 - u^2) \, du = \int (u^2 - 1) \, du = \frac{u^3}{3} - u + C \\\textbf{Back-substitute } u = \cos \theta \\\frac{\cos^3 \theta}{3} - \cos \theta + C \\\textbf{Evaluate from 0 to } \frac{\pi}{4} \\\left[ \frac{\cos^3 \theta}{3} - \cos \theta \right]_0^{\frac{\pi}{4}} = \left( \frac{\cos^3 \left( \frac{\pi}{4} \right)}{3} - \cos \left( \frac{\pi}{4} \right) \right) - \left( \frac{\cos^3 0}{3} - \cos 0 \right) \\= \left( \frac{\left( \frac{\sqrt{2}}{2} \right)^3}{3} - \frac{\sqrt{2}}{2} \right) - \left( \frac{1^3}{3} - 1 \right) \\= \left( \frac{\sqrt{2}^3}{3 \cdot 2^3} - \frac{\sqrt{2}}{2} \right) - \left( \frac{1}{3} - 1 \right) \\= \frac{\sqrt{2}}{12} - \frac{\sqrt{2}}{2} + \frac{2}{3} \\= \frac{-5\sqrt{2}}{12} + \frac{2}{3} \\ = \frac{2}{3} - \frac{5}{6\sqrt2}$$​​