# Thread: Another definite integral problem

1. ## Another definite integral problem

The question:

$\displaystyle \int_{0}^{\pi} \frac{sinx}{5 + 2cosx} dx$

My attempt:

Let u = 5 + 2cosx

$\displaystyle \frac{du}{dx} = -2sinx$

du = -2sinx dx

$\displaystyle -\frac{1}{2} \int_{0}^{\pi} \frac{1}{u} du$

$\displaystyle x = 0 \rightarrow u = 7$
$\displaystyle x = \pi \rightarrow u = 3$

$\displaystyle -\frac{1}{2} [log(u)]_{7}^{3}$

$\displaystyle -\frac{1}{2} [log(3) - log(7)]$

$\displaystyle -\frac{1}{2} [log(3/7)]$

$\displaystyle [log(3/7)^{-1/2}]$

$\displaystyle [log(\frac{\sqrt{7}}{\sqrt{3}})]$

However, the answer in my text is:

$\displaystyle \frac{1}{2}(ln(7) - ln(3))$

What am I doing wrong? Am I simplifying too far? Thanks.

2. Originally Posted by Glitch
$\displaystyle -\frac{1}{2} [log(3) - log(7)]$

....

What am I doing wrong? Am I simplifying too far? Thanks.
I think you are

$\displaystyle -\frac{1}{2} [\log(3) - \log(7)] =\frac{1}{2}(\log(7)-\log(3))$

3. $\displaystyle log(\frac{\sqrt{7}}{\sqrt{3}}) = log(\sqrt{\frac{7}{3}}) = log(\frac{7}{3})^{1/2} = \frac{1}{2}[log(7)-log(3)]$

$\displaystyle \therefore \left[log(\frac{\sqrt{7}}{\sqrt{3}})\right]$ is the same as $\displaystyle \frac{1}{2}(log(7) - log(3))$

4. Thank you guys.