## University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson

# Chapter 8 - Section 8.2 - Trigonometric Integrals - Exercises - Page 434: 20

#### Answer

$$\int^\pi_08\sin^4y\cos^2ydy=\frac{\pi}{2}$$

#### Work Step by Step

$$A=\int^\pi_08\sin^4y\cos^2ydy$$ $$A=\int^\pi_08(\sin^2y)^2\cos^2ydy$$ This is an example of Case 3, so we need to rewrite $$\sin^2y=\frac{1-\cos2y}{2}$$ $$\cos^2y=\frac{1+\cos2y}{2}$$ Therefore, $$A=\int^\pi_08\Big(\frac{1-\cos2y}{2}\Big)^2\Big(\frac{1+\cos2y}{2}\Big)dy$$ $$A=\int^\pi_08\frac{(1-\cos2y)^2}{4}\frac{1+\cos2y}{2}dy$$ $$A=\int^\pi_0(1-\cos2y)^2(1+\cos2y)dy$$ $$A=\int^\pi_0(1-2\cos2y+\cos^22y)(1+\cos2y)dy$$ $$A=\int^\pi_0(1-\cos2y-\cos^22y+\cos^32y)dy$$ $$A=\Big(y-\frac{\sin2y}{2}\Big)\Big]^\pi_0-\int^\pi_0(\cos^22y-\cos^32y)dy$$ $$A=\Big(\pi-\frac{\sin2\pi}{2}\Big)-\int^\pi_0(\cos^22y-\cos^32y)dy$$ $$A=\pi-\int^\pi_0(\cos^22y-\cos^32y)dy$$ - For $\int^\pi_0\cos^22ydy$: This is Case 3, so $$\int^\pi_0\cos^22ydy=\int^\pi_0\frac{1+\cos4y}{2}dy$$ $$=\frac{y}{2}+\frac{\sin4y}{8}\Big]^\pi_0$$ $$=\frac{\pi}{2}+\frac{\sin4\pi}{8}-0=\frac{\pi}{2}$$ - For $\int^\pi_0\cos^32ydy$: This is Case 2, so $$\int^\pi_0\cos^32ydy=\frac{1}{2}\int^{\pi}_0\cos^32yd(2y)$$ $$=\frac{1}{2}\int^{\pi}_0\cos^22y\cos2yd(2y)$$ $$=\frac{1}{2}\int^{\pi}_0(1-\sin^22y)d(\sin2y)$$ Take $u=\sin2y$ For $y=0$, $u=\sin0=0$ and for $y=\pi$, $u=\sin2\pi=0$ Therefore, $$\int^\pi_0\cos^32ydy=\frac{1}{2}\int^0_0(1-u^2)du=0$$ So, overall, $$A=\pi-\Big(\frac{\pi}{2}-0\Big)=\frac{\pi}{2}$$

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.