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I have read these days a nice way for integrating $\cos^3x$:

First differentiate:

$$f=\cos^3(x)$$$$f'=-3\cos^2(x)\sin(x)$$$$f''=6\cos(x)\sin^2(x)-3\cos^3(x)$$$$f''=6\cos(x)(1-\cos^2(x))-3f$$$$f''=6\cos(x)-6\cos^3(x)-3f$$$$f''=6\cos(x)-9f$$

Then integrate: $$f'= 6\sin(x) -9\int f$$ Therefore,

$$\int f=\frac 23 \sin(x) - \frac {f'} 9$$

Do you know any other way to calculate easily this integral?

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    $\begingroup$ The usual way is writing $\cos^3(x) = (1-\sin^2(x))\cos(x)$ and doing the obvious substitution. $\endgroup$ – AlohaSine Aug 30 '17 at 0:54
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    $\begingroup$ For common functions, you may use \function. Here, you may use \cos and \sin. Powers should not be done in unicode. For example, you should replace ² with ^2. $\endgroup$ – Simply Beautiful Art Aug 30 '17 at 0:56
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    $\begingroup$ Related: Evaluating $\int P(\sin x, \cos x) \text{d}x$ $\endgroup$ – Simply Beautiful Art Aug 30 '17 at 1:09
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    $\begingroup$ Given you the OP wants a list of techniques, a lot of them will be overkill. @SimplyBeautifulArt :) $\endgroup$ – Thomas Andrews Aug 30 '17 at 1:18
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    $\begingroup$ This is very elegant. (+1) for the approach. $\endgroup$ – BAI Aug 30 '17 at 4:09
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Maybe you'll like this method: \begin{eqnarray} \int\cos^3xdx&=&\int\cos^2xd\sin x\\ &=&\int(1-\sin^2x)d\sin x\\ &=&\sin x-\frac13\sin^3x+C. \end{eqnarray}

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  • $\begingroup$ Thanks really nice way to get it..I love it when you put your function under the d... $\endgroup$ – Aryadeva Aug 30 '17 at 3:13
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    $\begingroup$ @Isham Me too, they seem to be deemphasising this in college, these days. And +1. $\endgroup$ – Rene Schipperus Aug 30 '17 at 5:20
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    $\begingroup$ @Isham, you are welcome. $\endgroup$ – xpaul Aug 30 '17 at 15:55
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Euler's approach:

\begin{align}\int\cos^3(x)~dx&=\int\left[\frac{e^{ix}+e^{-ix}}2\right]^3~dx\\&=\int\frac{e^{3ix}+3e^{ix}+3e^{-ix}+e^{-3ix}}8~dx\end{align}

At this point, either integrate directly, and then notice the resulting sine functions, or notice that

$$\frac{e^{3ix}+e^{-3ix}}8=\frac14\cos(3x)\\\frac{3e^{ix}+3e^{-ix}}8=\frac34\cos(x)$$

And thus the rest is simple.

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Use that $\cos 3x = 4\cos^3 x - 3\cos x$ to get:

$$\int \cos^3 x \,dx = \frac{1}{4}\int\left(\cos 3x +3\cos x\right)\,dx$$

and the right side is easy to compute as $\frac{1}{12}\sin 3x +\frac{3}{4}\sin x$.


My original answer included a second approach, but I had the wrong formula for $dx$ in the substitution.

Use the tangent-half-angle substitution, $t=\tan(x/2)$ so $dx = \frac{2}{1+t^2}\,dt$ (corrected) and $\cos(x)=\frac{1-t^2}{1+t^2}$.

This reduces to:

$$\int \frac{2(1-t^2)^3}{(1+t^2)^4}\,dt$$

But this is a bit harder to integrate than the nice formula I had before. You might still solve this if you rewrite $1-t^2=2-(1+t^2)$ and expand, and then compute:

$$\int \frac{dt}{\left(1+t^2\right)^k}$$ for $k=1,2,3,4.$

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  • $\begingroup$ $\int\frac{(2-u)^3}{u^4}~du$ does not require partial fractions. $\endgroup$ – Simply Beautiful Art Aug 30 '17 at 1:20
  • $\begingroup$ @SimplyBeautifulArt Yeah, I had the wrong formula initially, but when I fixed, forgot to delete last line. $\endgroup$ – Thomas Andrews Aug 30 '17 at 1:20
  • $\begingroup$ But the fact that we end up with a $\ln\sec^2(x/2)$ in the formula seems a bit odd, so I'm wondering if I got that right $\endgroup$ – Thomas Andrews Aug 30 '17 at 1:22
  • $\begingroup$ It looks right... anyways, I have to head to bed. :-) $\endgroup$ – Simply Beautiful Art Aug 30 '17 at 1:24
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\begin{eqnarray*} \int \cos^3(x) dx &=& \int (1-\sin^2(x)) \cos(x) dx = \int \cos(x) dx - \int \sin^2(x) \cos(x) dx \\ \int \cos(x) dx &=& \sin(x) + C_1 \end{eqnarray*} Now, we need to integrate $\sin^2(x) \cos(x)$. We do this with the substitution $u = \sin(x)$. \begin{eqnarray*} \int \sin^2(x) \cos(x) dx &=& \int u^2 du = \frac{u^3}{3} + C_1 \\ \int \sin^2(x) \cos(x) dx &=& \frac{\sin^3(x)}{3} + C_1 \\ \int \cos^3(x) dx &=& \sin(x) - \frac{\sin^3(x)}{3} + C \\ \end{eqnarray*}

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A quite different approach would be to use integration by parts. Rewriting $\cos^3x$ as $(\cos^2x)(\cos x)$ and choose $f=\cos^2x$ and $g'=\cos x$, then integration by parts gives $\int{\cos^3}dx=\sin x\cos^2x+2\int{\cos x\sin^2x}dx$. When you replace $\sin^2x=1-\cos^2x$ you get another $\cos^3x$ term behind an integral sign. This way you can solve an equation in terms of $\int{\cos^3x}dx$.

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    $\begingroup$ Great or you can consider the second integral as the derivative of sin³(x) ..integrating a derivative...Since sin³x'= 3sin²cosx...you can deduce or do it your way with another cos³x. $\endgroup$ – Aryadeva Aug 30 '17 at 3:11
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Simpson approach...

$$ g =\int \cos^3(x)dx = \int \cos(x) \cos^2(x)dx$$

$$ g = \frac 12\int \cos(x) (\cos(2x)+1)dx$$

$$ 2 g = \int \cos(x) \cos(2x)dx + \int \cos(x) dx$$

$$ 2 g = \int \frac 12 (\cos(x) + \cos(3x))dx + \sin (x) $$

$$ g = \frac 1 {12} \sin(3x) + \frac 34 \sin (x) + k$$

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