Integration question (Log)

[lilshai]

Hello,
I have this equation:

d(cos x +/- isinx)/(cos x +/- isinx) = +/- idx

which yields after anti-differentiation,

ln(cos x +/- isinx) = +/- ix + c.

My question is, how did they get the left-hand side?

I know that the integral of 1/x is ln x, but what happens to the numerator (cos x +/- isinx) in the original equation?

 

[galactus]

It appears they're using the fact that \(\displaystyle e^{\pm\(ix)}=cos(x)\pm{isin(x)}\)


\(\displaystyle ln(cos(x)\pm{isin(x)})=ln(e^{\pm(ix)})=\pm{ix}\)

 

[pka]

I am not sure that I really understand the notation in your question.
Lets leave aside the +/− so that \(\displaystyle \L
d(\cos (x) + i\sin (x)) = ( - \sin (x) + i\cos (x))dx\).
Then \(\displaystyle \L
\frac{{d(\cos (x) + i\sin (x))}}{{\cos (x) + i\sin (x)}} = \frac{{( - \sin (x) + i\cos (x))dx}}{{\cos (x) + i\sin (x)}}\left( {\frac{{\cos (x) - i\sin (x)}}{{\cos (x) - i\sin (x)}}} \right) = \frac{{idx}}{1}\).

That gives you the complex algebra of how the two sides are equal to begin with.

The second part of question can best be done with an example.

If \(\displaystyle \L
f(x) = \ln (x\sin (x))\quad \mbox{then} \quad f'(x) = \frac{{\sin (x) + x\cos (x)}}{{x\sin (x)}}\).

Therefore, \(\displaystyle \L
\int {\frac{{\sin (x) + x\cos (x)}}{{x\sin (x)}}dx = \ln (x\sin (x)) + c}\).

 

[lilshai]

ok...I understand how they get ln(cos x (+/-) i*sin x) = (+/-)idx + c now.

The next step after that line is cos x (+/-) i*sinx = Ke^(+/-)ix

It appears that they took the natural log of both sides to get cos x (+/-) i*sinx for the left-hand side, but how do they go from (+/-)idx + c to Ke^(+/-)ix for the right-hand side?

Thanks.

 

[galactus]

ok...I understand how they get ln(cos x (+/-) i*sin x) = (+/-)idx + c now.

The next step after that line is cos x (+/-) i*sinx = Ke^(+/-)ix

It appears that they took the natural log of both sides to get cos x (+/-) i*sinx for the left-hand side, but how do they go from (+/-)idx + c to Ke^(+/-)ix for the right-hand side?

Thanks.

They took e to both sides:

I'm going to use 'plus' only, fro simplicity and less typing.

\(\displaystyle e^{ln(cos(x)+isin(x)}=e^{ix+c}\)

\(\displaystyle cos(x)+isin(x)=e^{ix}e^{c}\)

Let \(\displaystyle e^{c}=K\), a constant

\(\displaystyle cos(x)+isin(x)=Ke^{ix}\)

 

[pka]

galactus, that is an excellent explanation of that point about K.

 

[galactus]

Thank you, pka.

 

[lilshai]

They took e to both sides:

I'm going to use 'plus' only, fro simplicity and less typing.

\(\displaystyle e^{ln(cos(x)+isin(x)}=e^{ix+c}\)

\(\displaystyle cos(x)+isin(x)=e^{ix}e^{c}\)

Let \(\displaystyle e^{c}=K\), a constant

\(\displaystyle cos(x)+isin(x)=Ke^{ix}\)

oh ok, that makes sense. Thank you!

 

[lilshai]

galactus, that is an excellent explanation of that point about K.

I agree. Thanks for your help as well.