- #1
karush
Gold Member
MHB
- 3,269
- 5
$\large{67}$
$$\displaystyle
I=\int \frac{\sqrt{1-x^2 }}{x} \,dx
= \ln\left({\frac{\sqrt{1-x^2 }}{\left| x \right|}}\right)
+\sqrt{1-{x}^{2}}+C$$
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
$$\displaystyle
I=\int \frac{\sqrt{1-x^2 }}{x} \,dx
= \ln\left({\frac{\sqrt{1-x^2 }}{\left| x \right|}}\right)
+\sqrt{1-{x}^{2}}+C$$
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