Homework Help: Integration problem 2

1. Feb 20, 2013

PhizKid

1. The problem statement, all variables and given/known data
$\int \frac{sin2x}{1 + cos^{2}x} \textrm{ } dx$

2. Relevant equations

3. The attempt at a solution
$\int \frac{sin2x}{1 + cos^{2}x} \textrm{ } dx \\\\ \int \frac{2sinxcosx}{1 + cos^{2}x} \textrm{ } dx \\\\ u = cosx \\\\ -\int \frac{2u}{1 + u^{2}} \textrm{ } du \\\\ w = u^{2} \\\\ -\int \frac{1}{1 + w} \textrm{ } dw \\\\ q = 1 + w \\\\ -\int \frac{1}{q} \textrm{ } dq \\\\ -ln(q) = -ln(1 + w) = -ln(1 + u^{2}) = -ln(1 + cos^{2}x)$

I don't know what I did wrong

2. Feb 20, 2013

Curious3141

Your answer is correct, except for the omission of the constant of integration. Which makes all the difference in the world, really.

Here's an easier way to approach the integral. Note that the denominator is, in fact: $\frac{1}{2}(3 + \cos 2x)$. Can you go from there?

You'll find that the final answer you get using that method is different from your answer by only a constant ($\ln 2$), which means the answers are equivalent.

3. Feb 20, 2013

SammyS

Staff Emeritus
What's the derivative of a constant?

4. Feb 20, 2013

PhizKid

I don't see where that denominator is. The solution says it's: -ln(cos(2x) + 3) + C

5. Feb 20, 2013

SammyS

Staff Emeritus
What is $\displaystyle \ \ -\ln(1 + \cos^{2}x)-(-\ln(\cos(2x) + 3))\ ?$

6. Feb 20, 2013

PhizKid

Isn't it $ln(\frac{1 + cos^2x}{cos(2x) + 3})$? But what does $ln(cos(2x) + 3)$ have to do with anything?

7. Feb 20, 2013

Mute

Try plotting that function.

8. Feb 20, 2013

PhizKid

It looks like -ln(2)

9. Feb 20, 2013

SammyS

Staff Emeritus
I get ln(2) .

That's just a constant, so the two answers are equivalent.