Evaluate $\int_{-1}^{1} [ \frac{2}{3} x^3 + \frac{2}{3}(2-x^2)^{3/2}] dx $ Evaluate $$\int_{-1}^{1} \left[ \frac{2}{3} x^3 + \frac{2}{3}(2-x^2)^{3/2}\right] dx $$
My attempt :$$ \frac{2}{3} \left[\frac{x^4}{4}\right]_{x=-1}^{x=1} + \frac{2}{3} \left[\frac{(2-x^2)^\frac{-1}{2}}{-1/2}\right]_{x=-1}^{x=1}=0$$
Is  its True ?
 A: sadly, no.
the first part was completely true, although you could have used the property that $x^3$ is an odd function.
$$\frac{2}{3}\int^1_{-1}(2-x^2)^{3/2}dx$$
is a trigonometric integral.
your method ignored the internal $2-x^2$ and did an invalid substitution.
a better substetution is $x = \sqrt{2}\sin(\alpha),dx=\sqrt{2}\cos(\alpha)$ we have to change the bounds too, $\arcsin(\frac{1}{\sqrt{2}})=\frac{\pi}{4},\arcsin(-\frac{1}{\sqrt{2}})=-\frac{\pi}{4}$
$$\frac{2}{3}\int^{\frac{\pi}{4}}_{-\frac{\pi}{4}}4(1-\sin(\alpha)^2)^{3/2}\cos(\alpha)d\alpha$$
using everyone's favorite trigonometric identity $\cos(\alpha)^2+\sin(\alpha)^2=1$
$$\frac{8}{3}\int^{\frac{\pi}{4}}_{-\frac{\pi}{4}}(\cos(\alpha)^2)^{3/2}\cos(\alpha)d\alpha$$
$$\frac{8}{3}\int^{\frac{\pi}{4}}_{-\frac{\pi}{4}}(\cos(\alpha))^4d\alpha$$
using another trigonometric identity $cos(\alpha)^2=\frac{cos(2\alpha)+1}{2}$
$$\frac{8}{3}\int^{\frac{\pi}{4}}_{-\frac{\pi}{4}}\frac{(\cos(2\alpha)+1)^2}{4}d\alpha$$
$$\frac{2}{3}\int^{\frac{\pi}{4}}_{-\frac{\pi}{4}}[\cos(2\alpha)^2+2\cos(2\alpha)+1]d\alpha$$
using it again:
$$\frac{2}{3}\int^{\frac{\pi}{4}}_{-\frac{\pi}{4}}\frac{\cos(4\alpha)+1}{2}+2\cos(2\alpha)+1d\alpha$$
splitting the integral:
$$\frac{1}{3}\int^{\frac{\pi}{4}}_{-\frac{\pi}{4}}\cos(4\alpha)d\alpha+\frac{4}{3}\int^{\frac{\pi}{4}}_{-\frac{\pi}{4}}\cos(2\alpha)d\alpha+\int^{\frac{\pi}{4}}_{-\frac{\pi}{4}}1d\alpha$$
all of these integrals are very simple substitutions:
$$\frac{1}{12}[\sin(4\alpha)]^{\frac{\pi}{4}}_{-\frac{\pi}{4}}+\frac{2}{3}[\sin(2\alpha)]^{\frac{\pi}{4}}_{-\frac{\pi}{4}}+[\alpha]^{\frac{\pi}{4}}_{-\frac{\pi}{4}}$$
$$0 + \frac{4}{3}+\frac{\pi}{2}$$
or simply
$$\frac{4}{3}+\frac{\pi}{2}$$
A: Using property of even and odd functions, 
$$\int_{-1}^{1} \left(\frac{2}{3} x^3 + \frac{2}{3}(2-x^2)^{3/2}\right) dx $$
$$=\frac{4}{3}\int_{0}^{1}(2-x^2)^{3/2} dx $$
Let $x=\sqrt2\sin\theta\implies dx=\sqrt2\cos\theta d\theta$
$$=\frac{4}{3}\int_{0}^{1}(2-2\sin^2\theta)^{3/2} \sqrt2\cos\theta d\theta $$
$$=\frac{4}{3}\int_{0}^{\pi/4}4\cos^4\theta d\theta $$
$$=\frac{16}{3}\int_{0}^{\pi/4}\left(\frac{\cos4\theta+4\cos2\theta+3}{8}\right)d\theta $$
$$=\frac{2}{3}\int_{0}^{\pi/4}(\cos4\theta+4\cos2\theta+3)d\theta $$
$$=\frac{2}{3}\left(\frac{\sin4\theta}{4}+2\sin2\theta+3\theta\right)_0^{\pi/4} $$
$$=\frac{4}{3}+\frac{\pi}{2}$$
