Integral of \( \sin^3 x \)

Derivation and Evaluation

Evaluate the integral:

\[ \int \sin^3 x \, dx \]

Write the integrand \( \sin^3 x \) as a product of \( \sin^2 x \) and \( \sin x \):

\[ \int \sin^3 x \, dx = \int \sin^2 x \sin x \, dx \]

Use the trigonometric identity \( \sin^2 x = 1 - \cos^2 x \) to write the integral as:

\[ \int \sin^3 x \, dx = \int (1 - \cos^2 x) \sin x \, dx \]

Expand \( (1 - \cos^2 x)\sin x \) and write:

\[ \int \sin^3 x \, dx = \int \sin x \, dx - \int \cos^2 x \sin x \, dx \]

Use Integration by Substitution: let \( u = \cos x \), which gives \( \dfrac{du}{dx} = -\sin x \) or \( du = -\sin x \, dx \). Substituting this yields:

\[ \int \sin^3 x \, dx = \int \sin x \, dx + \int u^2 \, du \]

Use integral formulas to evaluate the integral:

\[ \int \sin^3 x \, dx = -\cos x + \dfrac{1}{3} u^3 + c \]

where \( c \) is the constant of integration.

Substitute back \( u = \cos x \) to find the final answer: