Please Help Evaluate the integral

[thatguy47]

\(\displaystyle \int_\)From 0 to (pi/4) of 4 tan^3xdx

I pulled out the 4 so it was 4 times int(0 > pi/4) of tan^3xdx

I'm tried doing u-substitution but I couldn't figure it out. Please help.

 

[soroban]

Hello, thatguy47!

$\displaystyle \int_0^{\frac{\pi}{4}\!}4\tan^3x\,dx$

$\displaystyle \text{We have: }\:4\!\int^{\frac{\pi}{4}}_0\! \tan^2x\,\tan x\,dx \:=\:4\!\int^{\frac{\pi}{4}}_0\!(\sec^2\!x - 1)\tan x\,dx$

. . $\displaystyle = \;4\!\int^{\frac{\pi}{4}}_0\!\bigg[\tan x\,\sec^2\!x - \tan x\bigg]\,dx \;=\;4\left[\int^{\frac{\pi}{4}}_0\!\tan x\,\sec^2\!x\,dx - \int^{\frac{\pi}{4}}_0\!\tan x\,dx\right]$

Got it?