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

$$\int 3\sin(x)\cos(x)\,dx=-\frac{3}{4}\cos(2x)+C.$$

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

Actually, we are to find the antiderivative of the given function, and all the functions you have found out to be potential answers are equally correct because any of them is just a constant shifted version of another. And these are not the only candidate to be the answers, we can find infinitely many more. When we differentiate all the results you have found out, we only get $3sinxcosx $ . And this is why when we try to find out the antiderivative of $3sinxcosx $ , we can achieve so many answers having a interesting relationship among them that, any two of them are just a constant value shifted from the other.

NB: Desmos is used to plot the graphs