Question

how do you integrate ln(2+sinx)?

Answer 1

im not sure some one can do it... :)
use wolfram to easier
http://www.wolframalpha.com/input/?i=int%28ln%282%2Bsin%28x%29%29%29dx

Answer 2

\[\int\limits \ln(2+\sin(x))dx\]
It might be tough.
IBP!
\[u=\ln(2+\sin(x)); dv=dx \implies u=\frac{\cos(x)}{2+\sin(x)}; v=x\]
\[\implies \ln(2+\sin(x))x-\int\limits \frac{x \cos(x)}{2+\sin(x)}dx\]
Now we can do integration by parts again. Make sure you DON'T integrate the cos(x)/(2+sin(x)) as it will just give you back the integral you started with.
\[u=\frac{\cos(x)}{2+\sin(x)}; dv=xdx \implies du=\frac{-\sin(x)(2+\sin(x))-\cos^2(x)}{(2+\sin(x))^2}=-\frac{2 \sin(x)+1}{(2+\sin(x))^2}\]
\[\alpha = \ln(2+\sin(x))x \implies \alpha - \left[ \frac{x^2 \cos(x)}{2(2+\sin(x))}+\int\limits \frac{x^2(\sin(x)+1)}{(2+\sin(x))^2}\right]\]
Maybe that doesn't help....

Answer 3

lol its fine.. thanks.... it just ive been thinkin for a day bout this and cant figure it out with the stuff my teacher taught me

Answer 4

I think you may be able to convert it over to the complex plane and do a contour integral. But I'm not extremely familiar with complex analysis integration techniques.

Answer 5

i never even knew those existed..... we just started integration in class..

Answer 6

i think if you get back the integral you started with you add and divide by 2 or some such trick
in any case it is not bad to get what you started with, you get an equation in that integral

Answer 7

on the other hand, looking at the wolf link, it is possible that there is not a nice closed form for this integral.
in other words, maybe you cannot do it, it just is what it is

Answer 8

Your first integration by parts created something. The secondary integrand is an odd function. Integration over a range symmetric around the y-axis is simple enough since the integral portion vanishes.

It's something!