# Integration Help (Quick if Possible!)

1. Oct 13, 2009

### Chewy0087

1. The problem statement, all variables and given/known data

$$\int sec^2 x tan x dx$$

2. Relevant equations

3. 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 $$\int tan x dx or \int u du$$

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

2. Oct 13, 2009

### Staff: Mentor

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.

3. Oct 13, 2009

### Chewy0087

i see now, much obliged sir!