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## Sine-cubed function

, 02:27, 27 August 2011
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|-
| [[antiderivative]] || $\frac{\cos^3x}{3} - \cos x + C$
|-
| important symmetries || [[odd function]]<br>[[half turn symmetry]] about all points of the form $(n\pi,0)$<br>[[mirror symmetry]] about all lines $x = n\pi + \pi/2$.
|}

==Identities==

We have the identity:

$\! \sin^3 x = \frac{3\sin x - \sin(3x)}{4}$

==Differentiation==

===First derivative===

To differentiate once, we use the [[chain rule for differentiation]]. Explicitly, we consider the function as the composite of the cube function and the sine function, so the cube function is the ''outer'' function and the sine function is the ''inner'' function.

We get:

$\frac{d}{dx}[(\sin x)^3] = 3(\sin x)^2 \cos x = 3\sin^2x \cos x$

<toggledisplay>Here's a more complete explanation: set $f$ as the function $t \mapsto t^3$ and $g$ as the function $x \mapsto \sin x$. Then, we have:

$\frac{d}{dx}(f(g(x))) = f'(g(x))g'(x)$

We have $f'(t) = 3t^2$, and $g'(x) = \cos x$, and plugging these into the expression gives the result above.</toggledisplay>

==Integration==

===First antiderivative: standard method===

We rewrite $\sin^3x = \sin^2x \sin x = (1 - \cos^2x)\sin x = (\cos^2x - 1)(-\sin x)$ and then do [[integration by u-substitution]] where $u = \cos x$. Explicitly:

$\int \sin^3 x \, dx = \int \sin^2x \sin x \, dx = \int (1 - \cos^2x) \sin x \, dx = \int (\cos^2x - 1) (-\sin x) \, dx$

Now put $u = \cos x$. We have $du/dx = - \sin x$, so we can replace $(-\sin x) \, dx$ by $du$, and we get:

$\int (u^2 - 1) \, du$

By polynomial integration, we get:

$\frac{u^3}{3} - u + C$

Plugging back $u = \cos x$, we get:

$\! \frac{\cos^3x}{3} - \cos x + C$.

Here, $C$ is an arbitrary real constant.

===First antiderivative: using triple angle formula===

An alternate method for integrating the function $\sin^3 x$ is to use the identity:

$\! \sin^3x = \frac{3\sin x - \sin(3x)}{4}$

We thus get:

$\! \int \sin^3x \, dx = \int \frac{3\sin x - \sin(3x)}{4} \, dx = \frac{-3}{4} \cos x + \frac{1}{12} \cos(3x) + C$

This answer looks superficially different from the other answer. However, using the identity $\cos(3x) = 4\cos^3x - 3\cos x$, we can verify that the antiderivatives are exactly the same.

===Repeated antidifferentiation===

The antiderivative of $\sin^3$ involves [[cos^3]] and [[cos]], both of which can be antidifferentiated, and this now involves [[sin^3]] and [[sin]]. We can thus antidifferentiate (i.e., integrate) the function any number of times, with the antiderivative expression alternating between a cubic function of sine and a cubic function of cosine.
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