# u substituion in integration

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• Dec 12th 2009, 10:09 AM
rainer
u substituion in integration
I cannot wrap my head around this u substitution business, please help.

For example...

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

$(u = \sin{x})$

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:

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

or

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

Thanks
• Dec 12th 2009, 10:55 AM
adkinsjr
Quote:

Originally Posted by rainer
I cannot wrap my head around this u substitution business, please help.

For example...

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

$(u = \sin{x})$

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

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

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

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

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

So the integral is:

$ln(sin(x))=\int cot(x)dx$
• Dec 12th 2009, 11:06 AM
adkinsjr
$ln(e^{ix})=\int\frac{1}{e^{ix}}du$

$u=e^{ix}\Rightarrow du=ie^{ix}dx$

$\int\frac{1}{e^{ix}}du=\int\frac{ie^{ix}dx}{e^{ix} }$

$=\int i dx$

$=ix$