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Finding anti-derivative of $f(x)= \sin^3 x \cos^2 x $

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

let $u = \cos x$

$\int -u^2 (1-u^2) du = \int -u^2 dx + \int u^4 dx = \frac{-u^3}{3} + \frac{u^5}{5} + C $

Therefore $ \frac{\cos^3 x}{3}+\frac{\cos^5 x}{5} + C$

Why am I wrong? My teacher said the answer is $ \frac{-\cos x}{8} - \frac{\cos (3x)}{48} + \frac{\cos (5x)}{80} $

user307640
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    You're not wrong. With the trigonometric functions, there's always identities can allow rewrites to happen. The two are the same. – Sean Roberson Oct 19 '22 at 01:47
  • @SeanRoberson Are they the same or a constant amount apart? For $x=0$ one seems to be is $\frac{8}{15}$ and the other $-\frac2{15}$ while for $x=\pi$ both signs reverse – Henry Oct 19 '22 at 01:53
  • Off by a constant. – Sean Roberson Oct 19 '22 at 01:56
  • I haven't checked the trigonometry between your teacher's answer and yours, but your integrals on the lower line should say $du$ not $dx$. – Suzu Hirose Oct 19 '22 at 01:56
  • You have a slight typo and should have the integral as $-\frac{\cos^3 x}{3}+\frac{\cos^5 x}{5} + C$. With that correction they are the same – Henry Oct 19 '22 at 01:58
  • A good way to check your result is correct or not is to compute its derivative and see whether it is equal to $\sin^3 x\cos^2x$ or not. – Taladris Oct 20 '22 at 08:52

3 Answers3

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Note that \begin{align} \cos^3 x &= \frac34\cos x +\frac14\cos 3x\\ \cos^5 x & = \frac58\cos x +\frac5{16}\cos 3x+\frac1{16}\cos 5x \end{align}

and \begin{align} &-\frac13\cos^3x+\frac15\cos^5x\\ =& -\frac13\left(\frac34\cos x +\frac14\cos 3x\right)+\frac15\left(\frac58\cos x +\frac5{16}\cos 3x+\frac1{16}\cos 5x\right)\\ =&-\frac18\cos x-\frac1{48}\cos 3x +\frac1{80}\cos5x \end{align} Thus, the two anti-derivatives are the same.

Ace
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To check the equivalence of different trigonometric forms, the simplest way is to use the following type of relation $$ \cos5\theta=\Re (e^{5i\theta}) $$ and the binomial theorem to derive relations between multiples of angles. For example, $$ \begin{align} \left({e^{i\theta}}\right)^5&=(\cos\theta+i\sin\theta)^5\\ &=\cos^5\theta+5i\cos^4\theta \sin\theta-10\cos^3\theta \sin^2\theta\\ &-10i\cos^2\theta\sin^3\theta+5\cos\theta\sin^4\theta+i\sin^5\theta\\ \end{align} $$ Discarding the complex part and writing $\sin^2=1-\cos^2$ gives us $$ \begin{align} \cos5\theta&=16\cos^5\theta-20\cos^3\theta+5\cos\theta \end{align} $$

Suzu Hirose
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First of all, Both answers $I=-\frac{\cos ^3 x}{3}+\frac{\cos ^5 x}{5}+C= \frac{1}{80} \cos 5 x-\frac{1}{48} \cos 3 x-\frac{1}{8} \cos x+C$ are correct, which can be confirmed by the following proof:

Let $z=\cos x+i\sin x$, then $\sin x= \frac{1}{2i}(z-\frac{1}{z})$ and $\cos x= \frac{1}{2}(z+\frac{1}{z})$.

Expressing $\sin^3x\cos^2x$ in terms of $z$ and then sine of multiples of $x$ gives $$ \begin{aligned} \sin ^3 x \cos ^2 x =&-\frac{1}{32 i}\left(z-\frac{1}{z}\right)^3\left(z+\frac{1}{z}\right)^2 \\ =&-\frac{1}{32 i}\left(z^2-\frac{1}{z^2}\right)^2\left(z-\frac{1}{z}\right) \\ =&-\frac{1}{32 i}\left[\left(z^5-\frac{1}{z^5}\right)-\left(z^3-\frac{1}{z^3}\right)-2\left(z-\frac{1}{z}\right)\right] \\ =&-\frac{1}{32 i}(2 i \sin 5 x-2 i \sin 3 x-4 i \sin x) \\ =&-\frac{1}{16} \sin 5 x+\frac{1}{16} \sin 3 x+\frac{1}{8}\sin x \end{aligned} $$ Integrating back yields $$ \boxed{\int \sin ^3 x \cos ^2 x d x=\frac{1}{80} \cos 5 x-\frac{1}{48} \cos 3 x-\frac{1}{8} \cos x+C} $$

Wish it helps!

Lai
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