Question

$\underset{-\pi /2}{\overset{\pi /2}{\int }}(x^3+x^2+{\tan }^{3}x)dx$

Answer 1

$I=\underset{-\pi /2}{\overset{\pi /2}{\int }}(x^3+x^2+{\tan }^{3}x)dx$

And, $f\left(x\right)=x^3+{\tan }^{3}x$, then

$f(-x)=(-x{)}^3+{\tan }^3(-x)=-x^3-{\tan }^{3}x=-f(x)$

Therefore, $f\left(x\right)$ is an odd function.

We know,

If $\varphi \left(x\right)$ is an odd function, then,

$\underset{-\pi /2}{\overset{\pi /2}{\int }}\varphi \left(x\right)dx=0$

And, if $\varphi \left(x\right)$ is enen function, then,

$\underset{-\pi /2}{\overset{\pi /2}{\int }}\varphi \left(x\right)dx=2\underset{0}{\overset{\pi /2}{\int }}\varphi \left(x\right)dx$

Therefore,

$I=\underset{-\pi /2}{\overset{\pi /2}{\int }}(x^3+{\tan }^{3}x)dx+\underset{-\pi /2}{\overset{\pi /2}{\int }}{x}^{2}dx$

$I=0+2\underset{0}{\overset{\pi /2}{\int }}{x}^{2}dx$

$I=\frac{2}{3}(\frac{\pi }{2}{)}^3=\frac{{\pi }^3}{12}$