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.