Question

Evaluate:
${\int }_{-\pi /2}^{\pi /2}(x^3+x\cos x+{\tan }^{5}x+1)dx$

Answer 1

${\int }_{\frac{-\pi }{2}}^{\frac{\pi }{2}}(x^3+x\cos x+{\tan }^{5}x+1)dx$

$={\left[\frac{x^4}{4}\right]}_{\frac{-\pi }{2}}^{\frac{\pi }{2}}+{\left[x\sin x\right]}_{\frac{-\pi }{2}}^{\frac{\pi }{2}}-{\int }_{\frac{-\pi }{2}}^{\frac{\pi }{2}}\sin xdx+{\int }_{\frac{-\pi }{2}}^{\frac{\pi }{2}}{\tan }^{3}x({\sec }^{2}x-1)dx+{\left[x\right]}_{\frac{-\pi }{2}}^{\frac{\pi }{2}}$

$=\frac{{\pi }^4}{64}-\frac{{\pi }^4}{64}+\frac{\pi }{2}-\frac{\pi }{2}-{[-\cos x]}_{\frac{-\pi }{2}}^{\frac{\pi }{2}}+{\int }_{\frac{-\pi }{2}}^{\frac{\pi }{2}}{\tan }^{3}x{\sec }^{2}xdx-{\int }_{\frac{-\pi }{2}}^{\frac{\pi }{2}}{\tan }^{3}xdx+\frac{\pi }{2}+\frac{\pi }{2}$

$=\pi +0+{\left[\frac{{\tan }^{4}x}{4}\right]}_{\frac{-\pi }{2}}^{\frac{\pi }{2}}-{\int }_{\frac{-\pi }{2}}^{\frac{\pi }{2}}\tan x{\sec }^{2}xdx-{\int }_{\frac{-\pi }{2}}^{\frac{\pi }{2}}\tan xdx$

$=\pi -{\left[\frac{{\tan }^{2}x}{2}\right]}_{\frac{-\pi }{2}}^{\frac{\pi }{2}}-{[-\log \left(\cos x\right)]}_{\frac{-\pi }{2}}^{\frac{\pi }{2}}$

$=\pi$