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

$$\int^{\pi}_{-\pi}(1-\cos^2t)^{3/2}dt=\frac{8}{3}$$
$$A=\int^{\pi}_{-\pi}(1-\cos^2t)^{3/2}dt$$ Use the identity: $$1-\cos^2t=\sin^2t$$ Therefore, $$A=\int^{\pi}_{-\pi}(\sin^2t)^{3/2}dt$$ $$A=\int^{\pi}_{-\pi}(|\sin t|)^3dt$$ As $\sin t\ge0$ on $[0,\pi]$ and $\lt0$ on $[-\pi,0)$, we have $$A=\int^\pi_0\sin^3tdt-\int^0_{-\pi}\sin^3tdt$$ Here is an example of Case 1, so we would rewrite $$\sin^3 tdt=\sin^2t\sin tdt=-(1-\cos^2t)d(\cos t)$$ $$A=-\int^\pi_0(1-\cos^2t)d(\cos t)+\int^0_{-\pi}(1-\cos^2t)d(\cos t)$$ $$A=\int^0_\pi(1-\cos^2t)d(\cos t)+\int^0_{-\pi}(1-\cos^2t)d(\cos t)$$ Set $u=\cos t$. - For $t=-\pi$, $u=-1$ - For $t=0$, $u=1$ - For $t=\pi$, $u=-1$ $$A=\int^1_{-1}(1-u^2)du+\int^1_{-1}(1-u^2)du=2\int^1_{-1}(1-u^2)du$$ $$A=2\Big(u-\frac{u^3}{3}\Big)\Big]^1_{-1}$$ $$A=2\Big[\Big(1-\frac{1}{3}\Big)-\Big(-1+\frac{1}{3}\Big)\Big]$$ $$A=2\Big(\frac{2}{3}+\frac{2}{3}\Big)=\frac{8}{3}$$