T thatguy47 Junior Member Joined Aug 11, 2008 Messages 69 Jan 12, 2009 #1 \(\displaystyle \int_\)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.
\(\displaystyle \int_\)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.
S soroban Elite Member Joined Jan 28, 2005 Messages 5,584 Jan 12, 2009 #2 Hello, thatguy47! \(\displaystyle \int_0^{\frac{\pi}{4}\!}4\tan^3x\,dx\) Click to expand... \(\displaystyle \text{We have: }\:4\!\int^{\frac{\pi}{4}}_0\! \tan^2x\,\tan x\,dx \:=\:4\!\int^{\frac{\pi}{4}}_0\!(\sec^2\!x - 1)\tan x\,dx\) . . \(\displaystyle = \;4\!\int^{\frac{\pi}{4}}_0\!\bigg[\tan x\,\sec^2\!x - \tan x\bigg]\,dx \;=\;4\left[\int^{\frac{\pi}{4}}_0\!\tan x\,\sec^2\!x\,dx - \int^{\frac{\pi}{4}}_0\!\tan x\,dx\right]\) Got it?
Hello, thatguy47! \(\displaystyle \int_0^{\frac{\pi}{4}\!}4\tan^3x\,dx\) Click to expand... \(\displaystyle \text{We have: }\:4\!\int^{\frac{\pi}{4}}_0\! \tan^2x\,\tan x\,dx \:=\:4\!\int^{\frac{\pi}{4}}_0\!(\sec^2\!x - 1)\tan x\,dx\) . . \(\displaystyle = \;4\!\int^{\frac{\pi}{4}}_0\!\bigg[\tan x\,\sec^2\!x - \tan x\bigg]\,dx \;=\;4\left[\int^{\frac{\pi}{4}}_0\!\tan x\,\sec^2\!x\,dx - \int^{\frac{\pi}{4}}_0\!\tan x\,dx\right]\) Got it?
