MHB LCC 205 8.4.12 sine substitution

[karush]

$\tiny{LCC \ \ 205 \ \ \ 8.4.12 \ \ sine \ \ substitution }$
$$\displaystyle
I=\int_{1/2}^{1} \frac{\sqrt{1-{x}^{2 }}}{{x}^{2 }} \ dx $$
Substitution
$$\displaystyle
x = \sin \left({u}\right)
\ \ \ dx=\cos\left({u}\right)\ du $$
Change of variables
$$\arcsin\left({1/2 }\right)=\frac{\pi}{6}=a
\ \ \ \arcsin\left({1}\right)=\frac{\pi}{2}=b$$
$$\displaystyle
I=\int_{a}^{b}
\frac{\sqrt{1-\sin^{2 }{u}}}{\sin^{2 }{u} }
\cos\left({u}\right) \ du
\implies
\int_{a}^{b}
\frac{\cos^{2 }{u}}{\sin^{2 }{u} } \ du
\implies\int_{a}^{b} \cot^2 \left({u}\right) \ du $$
By table reference
$$\displaystyle
I=-\cot{u}-u+C$$
Then
$$\displaystyle
I=\left[I\right]_{\pi/2}^{\pi/6} +C
=\sqrt{3}-\frac{\pi}{3}+C$$
No book answer, not sure of proper notation

$\tiny{\text{from math study group at Surf The Nations}}$

 

[Greg]

Should be $$\large{\left.I\right|^{\frac{\pi}{2}}_{\frac{\pi}{6}}}$$ and you don't need a constant of integration as it's a definite integral. Other than those two things it looks good. :)

 

[karush]

So then..
$\displaystyle
\large{\left.I\right|
^{\pi/2}_{\pi/6}}
=\sqrt{3}-\frac{\pi}{3} $