2023-03-14

How to find the integral of ${\mathrm{sin}}^{3}\left[x\right]dx$?

unieventos8l9

$\int {\mathrm{sin}}^{3}\left(x\right)dx=\int \mathrm{sin}\left(x\right)\left(1-{\mathrm{cos}}^{2}\left(x\right)\right)dx$
$=\int \mathrm{sin}\left(x\right)dx-\int \mathrm{sin}\left(x\right){\mathrm{cos}}^{2}\left(x\right)dx$

For the first integral:
$\int \mathrm{sin}\left(x\right)dx=-\mathrm{cos}\left(x\right)+C$

Using substitution, compute the second integral as follows:
Let $u=\mathrm{cos}\left(x\right)⇒du=-\mathrm{sin}\left(x\right)dx$
Then
$-\int \mathrm{sin}\left(x\right){\mathrm{cos}}^{2}\left(x\right)dx=\int {u}^{2}du$
$=\frac{{u}^{3}}{3}+C$
$=\frac{1}{3}{\mathrm{cos}}^{3}\left(x\right)+C$

Putting it all together, we get our final result:
$\int {\mathrm{sin}}^{3}\left(x\right)dx=\int \mathrm{sin}\left(x\right)dx-\int \mathrm{sin}\left(x\right){\mathrm{cos}}^{2}\left(x\right)dx$
$=-\mathrm{cos}\left(x\right)+\frac{1}{3}{\mathrm{cos}}^{3}\left(x\right)+C$

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