# Another definite integral problem

• Oct 29th 2010, 10:48 PM
Glitch
Another definite integral problem
The question:

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

My attempt:

Let u = 5 + 2cosx

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

du = -2sinx dx

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

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

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

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

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

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

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

However, the answer in my text is:

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

What am I doing wrong? Am I simplifying too far? Thanks.
• Oct 29th 2010, 10:59 PM
pickslides
Quote:

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

....

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

I think you are

$-\frac{1}{2} [\log(3) - \log(7)] =\frac{1}{2}(\log(7)-\log(3))$
• Oct 30th 2010, 12:04 AM
harish21
$log(\frac{\sqrt{7}}{\sqrt{3}}) = log(\sqrt{\frac{7}{3}}) = log(\frac{7}{3})^{1/2} = \frac{1}{2}[log(7)-log(3)]$

$\therefore \left[log(\frac{\sqrt{7}}{\sqrt{3}})\right]$ is the same as $\frac{1}{2}(log(7) - log(3))$
• Oct 30th 2010, 12:13 AM
Glitch
Thank you guys.