MHB What Is the Correct U-Substitution for This Integral?

[karush]

$\large{67}$
$$\displaystyle
I=\int \frac{\sqrt{1-x^2 }}{x} \,dx
= \ln\left({\frac{\sqrt{1-x^2 }}{\left| x \right|}}\right)
+\sqrt{1-{x}^{2}}+C$$
From a triangle this looks like a $\tan\left({\theta}\right)$ Integral

So wasn't sure what the u substitution should be

The answer was from TI-Inspire cx cas

 

[Greg]

Rewrite:

$$\int\dfrac{\sqrt{1-x^2}}{x}\,dx=\int\dfrac{1-x^2}{x\sqrt{1-x^2}}\,dx$$

Bust 'em up:

$$\int\dfrac{1}{x\sqrt{1-x^2}}\,dx-\int\dfrac{x}{\sqrt{1-x^2}}\,dx$$

The one on the left: use $x=\sin\theta$

The one on the right is equivalent to $\sqrt{1-x^2}+C$

Can you proceed?

 

[karush]

$$\displaystyle I=\int\dfrac{\sqrt{1-x^2}}{x}\,dx=\int\dfrac{1-x^2}{x\sqrt{1-x^2}}\,dx

=\int\dfrac{1}{x\sqrt{1-x^2}}\,dx-\int\dfrac{x}{\sqrt{1-x^2}}\,dx$$
solving
$$\displaystyle
\int\dfrac{1}{x\sqrt{1-x^2}}\,dx \\
x=\sin\theta \ \ \ \ dx=\cos\left({\theta}\right) \, d\theta \\
\int \frac{\cos\left({\theta}\right)}{\sin\left({\theta}\right)\cos\left({\theta}\right)} \, d\theta
= \int \frac{1}{\sin\left({\theta}\right)} \, d\theta
= \ln\left({\frac{\left| \sin\left({\theta}\right) \right|}{\cos\left({\theta}\right)+1}}\right)
$$
doing something ?

 

[Greg]

It's a "standard" integral:

$$\int\dfrac{1}{\sin\theta}\,d\theta=\int\csc\theta\,d\theta=-\ln|\cot\theta+\csc\theta|+C=-\ln\left(\dfrac{\sqrt{1-x^2}+1}{|x|}\right)+C$$

 

[karush]

TI gave something else... oh well