Integrating $\sin^3\left({x}\right)\cos\left({x}\right)dx$: Solution

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Discussion Overview

The discussion revolves around the integration of the function $\sin^3\left({x}\right)\cos\left({x}\right)$ with various approaches and methods being proposed. Participants explore different substitution techniques and transformations in the context of integral calculus.

Discussion Character

  • Mathematical reasoning
  • Exploratory
  • Technical explanation

Main Points Raised

  • One participant proposes the substitution $u=\cos\left({x}\right)$, leading to the integral $\int\left(u-u^3\right)du$, but expresses difficulty in reaching a solution.
  • Another participant suggests a different substitution, $u=\sin(x)$, which simplifies the integral to $\int {u}^{3}du$ and calculates it as $\frac{\sin^4\left({x}\right)}{4}+C$, noting it seems too straightforward.
  • A third participant provides a more complex approach by rewriting the integral using the identity for $\sin^2(x)$ and transforming it into a new integral involving $\sin(2x)$ and $\cos(2x)$, inviting others to propose alternate forms.
  • Some participants question the necessity of certain steps in the transformations, particularly regarding the manipulation of $\sin^2(x)$ and its derivative.

Areas of Agreement / Disagreement

There is no consensus on a single method for solving the integral, as multiple approaches are presented, and participants express varying opinions on the complexity and necessity of different steps.

Contextual Notes

Participants do not resolve the mathematical steps involved in their approaches, and there are assumptions made about the transformations that are not explicitly stated.

karush
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$\int\sin^3\left({x}\right)\cos\left({x}\right)dx$
$u=\cos\left({x}\right) du=\sin\left({x}\right)dx$
$\int\left(u-u^3\right)du$
I continued but didn't get the answer
 
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Since $$\frac{d}{dx}\left(\sin(x)\right)=\cos(x)$$, let:

$$u=\sin(x)\implies du=\cos(x)\,dx$$
 
So then...

$\int {u}^{3}du\implies \frac{u^4}{4}\implies\frac{\sin^4\left({x}\right)}{4}+C$

that's too easy!
 
Too easy? How about this then?

$\displaystyle \begin{align*} \int{\sin^3{(x)}\cos{(x)}\,\mathrm{d}x} &= \int{ \sin^2{(x)}\sin{(x)}\cos{(x)}\,\mathrm{d}x} \\ &= \int{ \frac{1}{2}\left[ 1 - \cos{(2x)} \right] \frac{1}{2}\sin{(2x)}\,\mathrm{d}x} \\ &= \frac{1}{8} \int{ \left[ 1 - \cos{(2x)} \right] \,2\sin{(2x)}\,\mathrm{d}x } \end{align*}$

Now let $\displaystyle \begin{align*} u = 1 - \cos{(2x)} \implies \mathrm{d}u = 2\sin{(2x)}\,\mathrm{d}x \end{align*}$...

What other alternate forms can you come up with?
 
Why doing the second step?

Since $$\dfrac{d}{dx}{{\sin }^{2}}x=2\cos (x)\sin(x)$$
 
Krizalid said:
Why doing the second step?

Since $$\dfrac{d}{dx}{{\sin }^{2}}x=2\cos (x)\sin(x)$$

Because I wanted it written in terms of (2x) :P
 

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