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## Homework Statement

[tex]\int sec^2 x tan x dx[/tex]

## Homework Equations

## The Attempt at a Solution

hey guys, sorry about this but i realised i can't do this question (in for tommorow)

I said, let u = tan x

so du/dx = sec^2 x

so dx/du =1 / sec^x

so putting that in I get [tex]\int tan x dx or \int u du[/tex]

which of course is [tex]\frac{u^2}{2}[/tex] = [tex]\frac{tan^2 x}{2}[/tex]

however according to wolfram alpha it's sec^2 x / 2 :O

http://www.wolframalpha.com/input/?i=int+sec^2+x+tan+x

thanks again in advance, would save me :O