# u substituion in integration

• Dec 12th 2009, 09:09 AM
rainer
u substituion in integration

For example...

$\displaystyle \ln(\sin{x})=\int\ \frac{1}{\sin{x}}du$

$\displaystyle (u = \sin{x})$

The integral is taken with respect to u, not x, which is totally worthless to me. I could care less about u. I'm all about x. So how do I get the integral to be dx ?

other examples causing me trouble:

$\displaystyle \ln(e^{ix})=\int\ \frac{1}{e^{ix}}du$ (I need to show how this results in ix)

or

$\displaystyle \ln(\sqrt{1-x^2})=\int\ \frac{1}{\sqrt{1-x^2}}du$

Thanks
• Dec 12th 2009, 09:55 AM
Quote:

Originally Posted by rainer

For example...

$\displaystyle \ln(\sin{x})=\int\ \frac{1}{\sin{x}}du$

$\displaystyle (u = \sin{x})$

The integral is taken with respect to u, not x, which is totally worthless to me. I could care less about u. I'm all about x. So how do I get the integral to be dx ?

Thanks

$\displaystyle ln(sin(x))=\int\frac{1}{sin(x)}du$

You have to find $\displaystyle du$ in terms of $\displaystyle x$

$\displaystyle u=sin(x)\Rightarrow du=cos(x)dx$

$\displaystyle \int\frac{du}{sin(x)}=\int\frac{cos(x)}{sin(x)}dx$

So the integral is:

$\displaystyle ln(sin(x))=\int cot(x)dx$
• Dec 12th 2009, 10:06 AM
$\displaystyle ln(e^{ix})=\int\frac{1}{e^{ix}}du$
$\displaystyle u=e^{ix}\Rightarrow du=ie^{ix}dx$
$\displaystyle \int\frac{1}{e^{ix}}du=\int\frac{ie^{ix}dx}{e^{ix} }$
$\displaystyle =\int i dx$
$\displaystyle =ix$