2
$\begingroup$

I am asked as a part of a question to integrate $$\int 2\,\sin^{2}{x}\cos{x}\,dx$$

I managed to integrate it using integration by inspection: $$\begin{align}\text{let } y&=\sin^3 x\\ \frac{dy}{dx}&=3\,\sin^2{x}\cos{x}\\ \text{so }\int 2\,\sin^{2}{x}\cos{x}\,dx&=\frac{2}{3}\sin^3x+c\end{align}$$

However, looking at my notebook the teacher did this: $$\int -\left(\frac{\cos{3x}-\cos{x}}{2}\right)$$ And arrived to this result: $$-\frac{1}{6}\sin{3x}+\frac{1}{2}\sin{x}+c$$

I'm pretty sure my answer is correct as well, but I'm curious to find out what how did do rewrite the question in a form we can integrate.

$\endgroup$
4
  • $\begingroup$ Let $ u = \sin x$. Also, this $\endgroup$ Jun 15, 2012 at 19:48
  • $\begingroup$ @TheChaz That's what he did... $\endgroup$ Jun 15, 2012 at 19:49
  • 2
    $\begingroup$ @David: No kidding. The authors of the two answers have interpreted the imparsible last sentence differently than I did. $\endgroup$ Jun 15, 2012 at 19:56
  • $\begingroup$ Your answer is absolutely correct, FWIW. $\endgroup$ Jun 15, 2012 at 19:56

6 Answers 6

6
$\begingroup$

$$2\sin^2(x)\cos(x) = (2\sin(x)\cos(x))\sin(x) = \sin(2x)\sin(x) = \frac{\cos(x) - \cos(3x)}{2}$$ The last step comes from : $$\cos(A - B) - \cos(A + B) = 2\sin(A)\sin(B)$$

$\endgroup$
4
$\begingroup$

$$2(\sin x)^2\cos x=(1-\cos 2x)\cos x=\cos x-\cos 2x\cos x=\cos x-\frac{1}{2}(\cos x +\cos 3x)$$ relying on $$\cos x \cos y =\frac{1}{2}(\cos(x-y)+\cos(x+y))$$

$\endgroup$
2
$\begingroup$

Try these identities: $$\sin(3x) = 3\sin(x) - 4\sin^3(x)$$ and $$\cos(3x) = 4\cos^3(x) - 3\cos(x)$$

$\endgroup$
1
$\begingroup$

Another natural approach is the substitution $u=\sin x$.

The path your instructor chose is less simple. We can rewrite $\sin^2 x$ as $1-\cos^2x$, so we are integrating $2\cos x-2\cos^3 x$. Now use the identity $\cos 3x=4\cos^3 x-3\cos x$ to conclude that $2\cos^3 x=\frac{1}{2}\left(\cos 3x+3\cos x\right)$.

Remark: The identity $\cos 3x=4\cos^3 x-3\cos x$ comes up occasionally, for example in a formula for solving certain classes of cubic equations. The same identity comes up when we are proving that the $60^\circ$ angle cannot be trisected by straightedge and compass.

$\endgroup$
1
$\begingroup$

Another way using a simple substitution:

$$I = \int 2\,\sin^{2}{x}\cos{x}\ \ dx$$

Let $u = \sin x, du = \cos x \ dx$

$$ I = 2\int u^2 \ du$$

$$I = \frac{2}{3} u^3$$

$$I = \frac{2}{3} \sin^3 x + C$$

$\endgroup$
1
$\begingroup$

I'd go direct, taking into account that $\,\displaystyle{\int x^2\,dx=\frac{1}{3}x^3 + C}\,$:$$\int 2\sin^2x\cos x\,dx=2\int\sin^2x(d\sin x)=\frac{2}{3}\sin^3x+C$$ This looks simpler to me and less messy than with trigonometric identities.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .