# 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.

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You can locate an several primitives of a function, they differ in a constant, but that does not mean that the results are wrong. –  Hiperion Sep 16 '13 at 5:50
Indefinite integration can be 'misleading' in this sense. –  copper.hat Sep 16 '13 at 5:52
A great question to ask students to see if they can come up with Zev's answer (or equivalent). –  nbubis Sep 16 '13 at 6:01
–  Jonas Meyer Sep 16 '13 at 6:07
When I was doing A-levels, I came across a similar example, but the two "different" answers to the integral involved respectively $\tan^2(x)$ and $\sec^2(x)$. Of course, these two expressions differ by a constant, but the explanation was non-obvious to me (and my dad) then, and I had to get my teacher to explain it. –  Hammerite Sep 16 '13 at 9:29

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

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

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Indeed this is a great demonstration of that's so important. –  David H Sep 16 '13 at 5:50
So, if is asked to solve the problem in the examination, which approach he should follow? His textbook contains only one of the answers. And as you can see, some graders may like only the conventional answers. No disrespect intended. –  Masroor Sep 16 '13 at 6:01
@MMA - All the answers are equally "wrong" since non of them contain a constant. When you let the constant C "swallow" the other constant term, all three give the same result. –  nbubis Sep 16 '13 at 6:02
@nbubis But his text book contains only one of the answers, (I can not get the book now), and given the prevailing circumstances, it will be difficult to modify the system. I can put down the answer which is the book in six/seven hours when the book comes back from school. –  Masroor Sep 16 '13 at 6:07
@MMA I sympathize with your situation, especially if your particular school has you feeling that cynical towards the system. But I have no way of predicting what which technique a grader would consider most appropriate, because there isn't one (and if the grader doesn't know it then your son knows more calculus than his teacher). He should learn all three methods. Solving integrals is like choosing a golf-club for a given shot. The best one is the one that gets the job done. –  David H Sep 16 '13 at 6:23

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

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