Differentiation concept problem

[quietrain]

Homework Statement

i am suddenly stuck when differentiating. are my concepts right?

if we integrate cosx/sinx, then since cosx is the differentiated form of sinx, then integrating d(sinx)/sinx = ln|sinx|

so now, if i have : integrate z/(1-z²)^1/2

since z is the differentiated form of the denominator(barring the minus sign), (1-z²)^1/2,
then integrating d((1-z²)^1/2)/(1-z²)^1/2,

shouldn't i get ln (1-z²)^1/2 ? but the answer is without the ln, and true enough when i try to differentiate the ln (1-z²)^1/2, i don't get back my original function.

so where did i go wrong?
thanks!

 

[SammyS]

It's definitely true that $\displaystyle \int\frac{d(\sqrt{1-z^2})}{\sqrt{1-z^2}}=\ln|\sqrt{1-z^2}|+C$

But notice that $\displaystyle d(\sqrt{1-z^2})=-\frac{z}{\sqrt{1-z^2}}dz\,.$

So, $\displaystyle \int\frac{d(\sqrt{1-z^2})}{\sqrt{1-z^2}}=-\int\frac{z\,dz}{\sqrt{1-z^2}}\,.$

 

[quietrain]

It's definitely true that $\displaystyle \int\frac{d(\sqrt{1-z^2})}{\sqrt{1-z^2}}=\ln|\sqrt{1-z^2}|+C$

But notice that $\displaystyle d(\sqrt{1-z^2})=-\frac{z}{\sqrt{1-z^2}}dz\,.$

So, $\displaystyle \int\frac{d(\sqrt{1-z^2})}{\sqrt{1-z^2}}=-\int\frac{z\,dz}{\sqrt{1-z^2}}\,.$

er i don't quite follow what you are saying?

the answer given for $\int\frac{z\,dz}{\sqrt{1-z^2}}\,.$ is just - (1-z²)^1/2

if i understand what you are saying, you are implying that the answer should be ln(1-z²)^1/2

?

 

[quietrain]

oh i see what you are trying to say.

just that i think you have a typo at the last expression

thanks!

 

[SammyS]

oh i see what you are trying to say.

just that i think you have a typo at the last expression

thanks!

Let's see:

$\displaystyle \frac{d}{dz}(\sqrt{1-z^2})=-\frac{z}{\sqrt{1-z^2}}\,.$

So, the last one seems right.