Integration Help (Quick if Possible)

[Chewy0087]

Homework Statement

\int sec^2 x tan x dx

Homework Equations

The Attempt at a Solution

hey guys, sorry about this but i realized 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 \int tan x dx or \int u du

which of course is \frac{u^2}{2} = \frac{tan^2 x}{2}

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

 

[Mark44]

Your answer is correct. Another substitution you can use is u = secx, du = secx tanx dx. With that substitution you have
\int u du~=~(1/2)u^2 + C~=~(1/2)sec^2x + C
The two answers seem to be different, and indeed they are. But they differ only by a constant, since (1/2)sec^2(x) = (1/2)tan^(x) + 1/2.

The upshot is that you are correct and wolfram alpha is also correct.

 

[Chewy0087]

i see now, much obliged sir!