How do you find the integral of $({sin}^{3} (x))$ $( sin^3(x))$ ?

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Answer 1 · 2016-04-21T14:58:03

$cos (x) - \frac{{cos}^{3} (x)}{3} + C$ $cos(x)-cos^3(x)/3+C$

We can integrate ${sin}^{3} (x)$ $sin^3(x)$ if we use the Pythagorean identity to rewrite it:

${sin}^{3} (x) = {sin}^{2} (x) sin (x) = (1 - {cos}^{2} (x)) sin (x)$ $sin^3(x)=sin^2(x)sin(x)=(1-cos^2(x))sin(x)$

Thus, we see that

$\int {sin}^{3} (x) d x = \int (1 - {cos}^{2} (x)) sin (x) d x$ $intsin^3(x)dx=int(1-cos^2(x))sin(x)dx$

We can now integrate through substitution, since we have both something and its derivative present.

Set $u = cos (x)$ $u=cos(x)$ , so that we also have $d u = sin (x) d x$ $du=sin(x)dx$ .

Substituting these both in, we see that

$intunderbrace((1-cos^2(x)))_(1-u^2)*underbrace(sin(x)dx)_(du)=int(1-u^2)du$

Integrating term by term, this gives us

$=u-u^3/3+C=cos(x)-cos^3(x)/3+C$

How do you find the integral of ( sin^3(x))(sin3(x))( sin^3(x))?