Answer to Question #301989 in Calculus for Arvie

Question

Evaluate the integral of sin³ y cos 2y dy from 0 to 𝜋/2.

Answer 1

$\begin{aligned} \int_{0}^{\frac{\pi}{2}} \sin^{3}x ~\cos 2x~ dx &= \int_{0}^{\frac{\pi}{2}} \dfrac{(3\sin x - \sin 3x)\cos 2x}{4}~ dx \quad \left(\sin^{3} x = \dfrac{3\sin x-\sin 3x}{4}\right)\\ &= \int_{0}^{\frac{\pi}{2}} \dfrac{3\sin x \cos 2x - \sin 3x \cos 2x}{4}~ dx\\ & \text{Using~} \sin \alpha \cos \beta = \dfrac{1}{2}(\sin(\alpha-\beta)+ \sin(\alpha+\beta)) \text{~we get}\\ &= \int_{0}^{\frac{\pi}{2}} \frac{1}{8}\left(3(\sin 3x - \sin x) - (\sin x + \sin 5x)\right)~ dx\\ &= \dfrac{1}{8} \int_{0}^{\frac{\pi}{2}} (3 \sin 3x - 3\sin x - \sin x - \sin 5x)~ dx\\ &= \dfrac{1}{8} \int_{0}^{\frac{\pi}{2}} (3 \sin 3x - 4\sin x - \sin 5x)~ dx\\ &= \dfrac{1}{8} \left(3 \left(\dfrac{-\cos 3x}{3}\right) - 4(-\cos x) - \left(-\dfrac{\cos 5x}{5}\right)\right)_{0}^{\frac{\pi}{2}}\\ &= \dfrac{1}{8} \left(-\cos 3x + 4\cos x + \dfrac{\cos 5x}{5}\right)_{0}^{\frac{\pi}{2}}\\ &= \dfrac{1}{8} \left((-\cos(\dfrac{3\pi}{2})+ 4\cos (\frac{\pi}{2}) + \dfrac{\cos (\frac{5\pi}{2})}{5})\right. - \\&\qquad \qquad\qquad\qquad\qquad\qquad\left.(-\cos 0+ 4\cos 0 + \dfrac{\cos 0}{5})\right)\\ &= \dfrac{1}{8} \left(0 - (-1+4+\dfrac{1}{5})\right)\quad(\text{Since~} \cos\frac{(2n + 1)\pi}{2} = 0 ~\forall n\in \N) \\ &= -\dfrac{2}{5} = -0.4 \end{aligned}$