Question

Prove that ${\int }_0^{\pi /4}2{\tan }^{3}xdx=1-\log 2$

Answer 1

$=2{\int }_0^{\frac{\pi }{4}}{\tan }^{3}xdx$

$=2{\int }_0^{\frac{\pi }{4}}{\tan }^{2}x\tan xdx$

$=2{\int }_0^{\frac{\pi }{4}}({\sec }^{2}x-1)\tan xdx$

$=2{\int }_0^{\frac{\pi }{4}}{\sec }^{2}x\tan xdx-2{\int }_0^{\frac{\pi }{4}}\tan xdx$

Let $t=\tan x\Rightarrow dt={\sec }^{2}xdx$

When $x=0\Rightarrow t=0$ and $x=\frac{\pi }{4}\Rightarrow t=1$

$=2{\int }_0^{1}tdt-{\left[\log \sec x\right]}_0^{\frac{\pi }{4}}$

$=2{\left[\frac{t^2}{2}\right]}_0^1-2(\log \sec \frac{\pi }{4}-\log \sec 0)$

$=[1-0]-2[\log \sqrt{2}-\log 1]$

$=1-\frac{2}{2}\log 2$

$=1-\log 2$

Hence proved.