# Chapter 8: Techniques of Integration - Section 8.3 - Trigonometric Integrals - Exercises 8.3 - Page 462: 32

$$\frac{8}{3}$$

#### Work Step by Step

We integrate as follows: \begin{align*} \int_{-\pi}^{\pi}\left(1-\cos ^{2} t\right)^{3 / 2} d t&=\int_{-\pi}^{\pi}\left(\sin ^{2} t\right)^{3 / 2} d t\\ &=\int_{-\pi}^{\pi}\left|\sin ^{3} t\right| d t\\ &=-\int_{-\pi}^{0} \sin ^{3} t d t+\int_{0}^{\pi} \sin ^{3} t d t\\ &=-\int_{-\pi}^{0}\left(1-\cos ^{2} t\right) \sin t d t+\int_{0}^{\pi}\left(1-\cos ^{2} t\right) \sin t d t\\ &=-\int_{-\pi}^{0} \sin t d t+\int_{-\pi}^{0} \cos ^{2} t \sin t d t+\int_{0}^{\pi} \sin t d t-\int_{0}^{\pi} \cos ^{2} t \sin t d t \$ \\ &=\left[\cos t-\frac{\cos ^{3} t}{3}\right]_{-\pi}^{0}+\left[-\cos t+\frac{\cos ^{3} t}{3}\right]_{0}^{\pi}\\ &=\left(1-\frac{1}{3}+1-\frac{1}{3}\right)+\left(1-\frac{1}{3}+1-\frac{1}{3}\right)\\ &=\frac{8}{3} \end{align*}

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.