Question

Integral [pi/4 to pi/3] tanx/[cos^2(x)*sqrt{6+tan^2(x)}]dx. I just don't know where to begin. I can't figure out any u-substition that would work.

Answer 1

@Loser66

Answer 2

Btw Welcome to Openstudy @Dnario :)

Answer 3

use latex to make it clear, please

Answer 4

@dan815 come here to play, friend

Answer 5

I have no idea how to use latex

Answer 6

\[\int\limits_{\pi/4}^{\pi/3}\frac{ \tan(x) }{ \cos^2(x)\times \sqrt{6+\tan^2 (x)}}dx\]

Answer 7

let u = tan x , so du = sec^2x dx, change limit

yours = tan.sec^2 / 6 + tan^2
by substitute, you have u/\sqrt ( 6 + u^2 ) du. Sub again

Answer 8

ohhl let dan helps