Question

Prove: $∫^{\frac{π}{4}}_0 2tan^3xdx=1-log2$

Answer 1

Let $I=∫^{\frac{π}{4}}_0 2tan^3xdx$
$I=2∫^{\frac{π}{4}}_0 tan^2x\,tanxdx=2∫^{\frac{π}{4}}_0(sec^2x-1)tanxdx$
$=2∫^{\frac{π}{4}}_0 sec^2x\,tanx\,dx-2∫^{\frac{π}{4}}_0 tanx\,dx$
$=2[\frac{tan^2x}{2}]^{\frac{π}{4}}_0+2[log\,cosx]^{\frac{π}{4}}_0$
$=1+2[logcos\frac{π}{4}-logcos0]$
$=1+2[log\frac{1}{\sqrt{2}}-log1]$
$=1-log2-log1=1-log2$
Hence,the given result is proved.