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.

All three answers are correct provided you add a constant to each one of those.
Because from the very definition of integration, it is the area under the curve, so it requires bounds to give a unique value.
You can't evaluate the value of an indefinite integral without including constant.
And I am sure that in the examination, your son won't be asked to evaluate the value of an integral without providing limits of integration or providing its value at some other point.
For instance, in question it may be mentioned that evaluate the value of expression at x=π/6 , given its value at x=0 is 1.
So in this case, all three answers will give the correct value i.e. 11/8
