# Thread: can someone confirm this

1. ## can someone confirm this

the integral from 0 to pi/2 of [(sinx)/(cos(x)+1)]

is -ln lcosx+1l + c

2. $\displaystyle \int_0^{\frac{\pi}{2}} \frac{ \sin x}{\cos x +1}~dx$

$\displaystyle = \left[ - \ln| \cos x+1| \right]_0^{\frac{\pi}{2}}$

$\displaystyle =- \ln 1+ \ln 2= \ln 2$

You forgot to put in your limits!

3. Originally Posted by Showcase_22
$\displaystyle \int_0^{\frac{\pi}{2}} \frac{ \sin x}{\cos x +1}~dx$

$\displaystyle = \left[ - \ln| \cos x+1| \right]_0^{\frac{\pi}{2}}$

$\displaystyle =- \ln 1+ \ln 2= \ln 2$

You forgot to put in your limits!
so its just ln2?

4. Yes! What you evaluated was the integral without limits (or the "indefinite integral").

When you have limits you can just drop the +c and start substituting.

5. Originally Posted by Showcase_22
Yes! What you evaluated was the integral without limits (or the "indefinite integral").

When you have limits you can just drop the +c and start substituting.
i dont understand how a -ln1+ln2=ln2. shouldnt it just be pos ln1?

6. Well $\displaystyle \ln 1=0$ so you can just get rid of it.

Alternatively, use the laws of logs:

$\displaystyle \ln 2 - \ln 1=\ln \frac{2}{1}= \ln 2$

7. Originally Posted by Showcase_22
Well $\displaystyle \ln 1=0$ so you can just get rid of it.

Alternatively, use the laws of logs:

$\displaystyle \ln 2 - \ln 1=\ln \frac{2}{1}= \ln 2$
thanks