# Homework Help: Constant of Integration?

1. Feb 21, 2015

### tree.lee

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

So, I have a trigonometric substitution integration problem. The working is rather hairy, but I've gotten to the point where you draw the triangle to express theta in terms of x. But that's where I'm stuck! I think I may be having trouble with the constant of integration, but I'm not sure!
2. Relevant equations
So...for the sides of the triangle I have:
Opposite: √[(x+½)2 - ¾]
Hypotenuse: x+½

My equation is tanθ-[½(ln|secθ+tanθ|)] and I want to express it in terms of x.

3. The attempt at a solution
So I just plug it in, Opposite/Adjacent and Hypotenuse/Adjacent but I'm getting it wrong.

I get √[(x+½)2 - ¾] / √¾ - ½ln|x + ½ + √[(x+½)2-¾]| / √(¾)
Which equals √(x2+x+1)/√¾ - ½ln|x+½+√(x2+x+1) + C

But it doesn't, the answer given is √(x2+x+1) - ½ln|x+½+√(x2+x+1), as in the only difference is the denominator for the first term. But I don't understand how it could be integrated into the integration constant, it's not a constant, it's dividing the variable x, no? Or is something else entirely wrong with it!?Any help would be greatly appreciated!

2. Feb 21, 2015

### Svein

Your own expressions give Hypotenuse/Adjacent = (x + ½)/√(¾) and Opposite/Adjacent = √[(x+½)2 - ¾]/√(¾). Common denominator, so add them. But - you have simply pushed √(¾) outside the ln(| |) expression, which you are not allowed to do.

3. Feb 21, 2015

### tree.lee

Oh. I was taught that I could do ½ln|x + ½ + √[(x+½)2-¾]| - ln√(¾) so that C1 = ln√(¾) + C. Is that wrong??? Oh, and it's also the first term that I"m having denominator troubles with. The tanθ! I believe I did add the secθ and tanθ with common denominators within the ln expression, as you said.

4. Feb 21, 2015

### Svein

Yes. But that is not what you wrote in your original post. And, by the way, don't forget the ½ when you move ln√(¾) outside.