What is the integral of #int ( sin^3(x))dx#?

Question

Paisley Johnson · Accepted Answer

#cos^3x/3-cosx+c# #intsinxsin^2xdx=intsinx(1-cos^2x)dx=intsinxdx-intsinxcos^2xdx=-cosx+cos^3x/3+c#

Cameron Alexander · Answer

undefined The integral of $ \int \sin^3(x) \, dx $ can be solved using trigonometric identities and integration by parts. The result is:

$$ \int \sin^3(x) \, dx = -\frac{\cos(x)\sin^2(x)}{2} + \frac{1}{3}\sin^3(x) + C $$

Where $ C $ is the constant of integration.

Axel Alvarez · Answer

undefined The integral of $ \int \sin^3(x) \, dx $ can be solved using trigonometric identities and integration by parts.

Using the trigonometric identity $ \sin^2(x) = 1 - \cos^2(x) $, we rewrite $ \sin^3(x) $ as $ \sin^2(x) \cdot \sin(x) $.

$$ \sin^3(x) = \sin^2(x) \cdot \sin(x) = (1 - \cos^2(x)) \cdot \sin(x) $$

Now, we can use a substitution $ u = \cos(x) $ and $ du = -\sin(x) \, dx $ to rewrite the integral in terms of $ u $.

$$ \int \sin^3(x) \, dx = \int (1 - u^2) \, (-du) $$

$$ = \int (u^2 - 1) \, du $$

Integrating term by term:

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

Substituting back $ u = \cos(x) $:

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

Therefore, the integral of $ \int \sin^3(x) \, dx $ is $ \frac{\cos^3(x)}{3} - \cos(x) + C $, where $ C $ is the constant of integration.