Different Ways of Integrating $3\sin x\cos x$ I am asking this question for my son who is in (equivalent) twelfth
grade and I failed to answer his query.
When he tries to integrate $3\sin x\cos x$, he finds that
this can be done in at least following three ways.
And these three ways do not produce equivalent results.
ONE
Let us assume, $\sin x = z$.
This gives,
\begin{align*}
  \cos x &= \frac{dz}{dx}\\
  \cos x dx &= dz
\end{align*}
So, we can write, 
\begin{align*}
  \int 3\sin x\cos x dx  &=3 \int zdz\\
  &=3 \frac{z^2}{2}\\
  &=\frac{3}{2} \sin^2 x\\
  &=\frac{3}{4}\times 2\sin^2 x\\
  &=\frac{3}{4} (1 -\cos 2x)\\
\end{align*}
TWO
Let us assume, $\cos x = z$.
This gives,
\begin{align*}
  -\sin x &= \frac{dz}{dx}\\
  \sin x dx &= -dz
\end{align*}
So, we can write, 
\begin{align*}
  \int 3\sin x\cos x dx  &=-3 \int zdz\\
  &=-3 \frac{z^2}{2}\\
  &=-\frac{3}{2} \cos^2 x\\
  &=-\frac{3}{4}\times 2\cos^2 x\\
  &=-\frac{3}{4} (1 +\cos 2x)\\
\end{align*}
THREE
\begin{align*}
  \int 3\sin x\cos x dx  &=\frac{3}{2}\int 2\sin x\cos x dx\\
  &=\frac{3}{2}\int \sin 2x dx\\
  &=-\frac{3}{2}\times\frac{1}{2} \cos 2x\\
  &=-\frac{3}{4} \cos 2x\\
\end{align*}
The results found in above three methods are not the same.
If we try a simple approach of evaluating the integration results at, $x = \frac{\pi}{6}$, we get as follows.
From the first one,
$\frac{3}{4} (1 -\cos 2x) = \frac{3}{4} (1 -\cos \frac{2\pi}{6})
= \frac{3}{4} (1 -\cos \frac{\pi}{3})
= \frac{3}{4} (1 - \frac{1}{2})
= \frac{3}{4}\times\frac{1}{2}
= \frac{3}{8}$
From the second one,
$-\frac{3}{4} (1 +\cos 2x) = -\frac{3}{4} (1 +\cos \frac{2\pi}{6})
= -\frac{3}{4} (1 +\cos \frac{\pi}{3})
= -\frac{3}{4} (1 + \frac{1}{2})
= -\frac{3}{4}\times\frac{3}{2}
= -\frac{9}{8}$ 
From the third one,
$-\frac{3}{4} \cos 2x=-\frac{3}{4} \cos \frac{2\pi}{6}
= -\frac{3}{4} \cos \frac{\pi}{3}
= -\frac{3}{4} \times \frac{1}{2}
= -\frac{3}{8} $
Clearly, we are getting some nonequivalent results. We have failed to find
the mistakes or explanations behind this. Your help will be appreciated.

You're forgetting that an indefinite integral must include a constant of integration; for any chosen constant $C$, we have that
$$\frac{d}{dx}\left(-\frac{3}{4}\cos(2x)+C\right)=3\sin(x)\cos(x),$$
and that is precisely the relationship captured by the statement that
$$\int 3\sin(x)\cos(x)\,dx=-\frac{3}{4}\cos(2x)+C.$$
