Question

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

Answer 1

Let $I={\int }_0^{\frac{\pi }{4}}2{\tan }^{3}xdx$
$I=2{\int }_0^{\frac{\pi }{4}}{\tan }^{2}x\tan xdx=2{\int }_0^{\frac{\pi }{4}}({\sec }^{2}x-1)\tan xdx$
$=2{\int }_0^{\frac{\pi }{4}}{\sec }^{2}x\tan xdx-2{\int }_0^{\frac{\pi }{4}}\tan xdx$
$=2{\left[\frac{{\tan }^{2}x}{2}\right]}_0^{\frac{\pi }{4}}+2[\log \cos x{]}_0^{\frac{\pi }{4}}$
$=1+2[\log \cos \frac{\pi }{4}-\log \cos 0]$
$=1+2[\log \frac{1}{\sqrt{2}}-\log 1]$
$=1-\log 2-\log 1=1-\log 2$
Hence, the given result is proved.