MHB Evaluate Integral $t_{1.11}$: $\cos^3$ to $\cos$

[karush]

$\tiny{t1.11}$
$\textsf{Evaluate the Integral}$
\begin{align*}\displaystyle
I_{11}&=\int \frac{\sin\sqrt{t}}{\sqrt{t\cos^3\sqrt{t}}}\, dt\\
&=\int\frac{\sin\sqrt{t}}{\sqrt{t}\cos^{3/2}\sqrt{t}}\, dt\\
u&=\cos\sqrt{t}\\
du&=-\dfrac{\sin\left(\sqrt{t}\right)}{2\sqrt{t}}\, dt\\
dt&=\frac{2\sqrt{t}}{\sin(\sqrt{t})}\, du\\
I_{11}&=\int\frac{\sin\sqrt{t}}{\sqrt{t}\cos^{3/2}\sqrt{t}}\, \frac{2\sqrt{t}}{\sin(\sqrt{t})} \, du\\
\textit{plug u in}&\\
&=2\int\frac{1}{u^{3/2}} du\\
&=\dfrac{4}{\sqrt{u}}+C\\
\textit{substitute back $\cos\sqrt{t}$ for u}&\\
I_{11}&=\dfrac{4}{\sqrt{\cos\left(\sqrt{t}\right)}}+C
\end{align*}
looking at an example I didn't understand why $\cos^3 x$ in the given Integral could not be simply reduced to just $\cos x$ by dropping the radical. I tried this but it didn't work.

 

[Prove It]

$\tiny{t1.11}$
$\textsf{Evaluate the Integral}$
\begin{align*}\displaystyle
I_{11}&=\int \frac{\sin\sqrt{t}}{\sqrt{t\cos^3\sqrt{t}}}\, dt\\
&=\int\frac{\sin\sqrt{t}}{\sqrt{t}\cos^{3/2}\sqrt{t}}\, dt\\
u&=\cos\sqrt{t}\\
du&=-\dfrac{\sin\left(\sqrt{t}\right)}{2\sqrt{t}}\, dt\\
dt&=\frac{2\sqrt{t}}{\sin(\sqrt{t})}\, du\\
I_{11}&=\int\frac{\sin\sqrt{t}}{\sqrt{t}\cos^{3/2}\sqrt{t}}\, \frac{2\sqrt{t}}{\sin(\sqrt{t})} \, du\\
\textit{plug u in}&\\
&=2\int\frac{1}{u^{3/2}} du\\
&=\dfrac{4}{\sqrt{u}}+C\\
\textit{substitute back $\cos\sqrt{t}$ for u}&\\
I_{11}&=\dfrac{4}{\sqrt{\cos\left(\sqrt{t}\right)}}+C
\end{align*}
looking at an example I didn't understand why $\cos^3 x$ in the given Integral could not be simply reduced to just $\cos x$ by dropping the radical. I tried this but it didn't work.

Maybe because a square root undoes a SQUARE, not a cube...