MHB 7.3.5 Integral with trig substitution

[karush]

$\textsf{Evaluate the integral}$
$$I=\displaystyle\int\frac{x^2}{\sqrt{9-x^2}}$$
$\textit{from the common Integrals Table we have}$
$$\displaystyle I=\int\frac{u^2}{\sqrt{u^2-a^2}} \, du
=\frac{u}{2}\sqrt{u^2-a^2}+\frac{a^2}{2}
\ln\left|u+\sqrt{u^2-a^2}\right|+C$$
$\textit{Thus if $u=x$ and $a=3$ we have}$
$$I_{44}=\int\frac{x^2}{\sqrt{x^2-3^2}} \, dx
=\frac{x}{2}\sqrt{x^2-3^2}+\frac{3^2}{2}
\ln\left|x+\sqrt{x^2-3^2}\right|+C$$
$\textit{or simplified}$
$$I_{44}=\int\frac{9}{\sqrt{x^2-9}} \, dx
=\frac{x}{2}\sqrt{x^2-9}+\frac{9}{2}
\ln\left|x+\sqrt{x^2-9}\right|+C$$

$\textit{so by using trig substitution to derive this we have}\\$
$\textit{$x=3\sin{u}\quad dx=3\cos{u} \, du $}$
\begin{align*}\displaystyle
&=\int\frac{9\sin^2{u}}{\sqrt{9-9\sin^2{u}}} \,
3\cos{u} du \\
&=9\int\frac{\sin^2{u}}{3\sqrt{1-\sin^2{u}}}
\, 3\cos{u} du
=9\int\sin^2{u} \, du\\
% &=9\int\frac{1 - cos^2{u}}{cos{u}} \, du
% = 9\left[\int\frac{1}{cos{u}}\, du - \int\frac{cos^2u}{cos u}\right]\\
\end{align*}ok how is this going to become what we see in the table equation ?

 

[LCKurtz]

$\textsf{Evaluate the integral}$
$$I=\displaystyle\int\frac{x^2}{\sqrt{9-x^2}}$$
$\textit{from the common Integrals Table we have}$
$$\displaystyle I=\int\frac{u^2}{\sqrt{u^2-a^2}} \, du
=\frac{u}{2}\sqrt{u^2-a^2}+\frac{a^2}{2}
\ln\left|u+\sqrt{u^2-a^2}\right|+C$$
$\textit{Thus if $u=x$ and $a=3$ we have}$
$$I_{44}=\int\frac{x^2}{\sqrt{x^2-3^2}} \, dx
=\frac{x}{2}\sqrt{x^2-3^2}+\frac{3^2}{2}
\ln\left|x+\sqrt{x^2-3^2}\right|+C$$
$\textit{or simplified}$
$$I_{44}=\int\frac{9}{\sqrt{x^2-9}} \, dx
=\frac{x}{2}\sqrt{x^2-9}+\frac{9}{2}
\ln\left|x+\sqrt{x^2-9}\right|+C$$

$\textit{so by using trig substitution to derive this we have}\\$
$\textit{$x=3\sin{u}\quad dx=3\cos{u} \, du $}$
\begin{align*}\displaystyle
&=\int\frac{9\sin^2{u}}{\sqrt{9-9\sin^2{u}}} \,
3\cos{u} du \\
&=9\int\frac{\sin^2{u}}{3\sqrt{1-\sin^2{u}}}
\, 3\cos{u} du
=9\int\sin^2{u} \, du\\
% &=9\int\frac{1 - cos^2{u}}{cos{u}} \, du
% = 9\left[\int\frac{1}{cos{u}}\, du - \int\frac{cos^2u}{cos u}\right]\\
\end{align*}ok how is this going to become what we see in the table equation ?

I don't think it will, but it might become$$
-\frac x 2\sqrt{9-x^2}+\frac 9 2 \arcsin\left( \frac x 3 \right )+C$$which I believe is the correct answer.