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

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Answer 1 · 2016-11-09T11:09:38

$= \frac{{cos}^{3} x}{3} - cos x + C$ $=(cos^3x)/3-cosx+C$ $" "$ C is a constant.

$\int {(sin x)}^{3} d x$ $int(sinx)^3dx$

$\int sin x {(sin x)}^{2} d x$ $intsinx(sinx)^2dx$

Let $u = cos x$ $color(red)(u = cosx" " )$ then $d u = - sin x d x \Rightarrow sin x d x = - d u$ $" "du=-sinxdx" "rArr color(red)(sinxdx = -du)$

Knowing the trigonometric identity:

${cos}^{2} x + {sin}^{2} x = 1$ $cos^2x + sin^2x =1$

${sin}^{2} x = 1 - {cos}^{2} x$ $sin^2x=1 - cos^2x$

$\int (- d u) {sin}^{2} x$ $int(-du)sin^2x$

$= \int - (1 - {cos}^{2} x) d u$ $=int-(1-cos^2x)du$

$= \int - (1 - u^{2}) d u$ $=int-(1-u^2)du$

$= \int u^{2} - 1 d u$ $=intu^2-1du$

$= \int u^{2} d u - \int 1 d u$ $=intu^2du-int1du$

$= \frac{u^{3}}{3} - u + C$ $=u^3/3-u+C$

Substituting $u = cos x$ $color(red)(u=cosx)$

$= \frac{{cos}^{3} x}{3} - cos x + C$ $=(cos^3x)/3-cosx+C$ $" "$ C is a constant.

How do you find the antiderivative of (sinx)3(sinx)^3?