Question

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

Hint:

Express odd powers of sine or cosine using the Pythagorean identity and substitute to simplify integration.

Answer 1

Rewrite \(\sin^3 x\) as: \[ \sin^3 x = \sin x \cdot \sin^2 x = \sin x (1 - \cos^2 x). \] So, \[ \int \sin^3 x \, dx = \int \sin x (1 - \cos^2 x) \, dx = \int \sin x \, dx - \int \sin x \cos^2 x \, dx. \] Let \(t = \cos x \implies dt = -\sin x \, dx\). Then, \[ \int \sin x \cos^2 x \, dx = -\int t^2 \, dt = -\frac{t^3}{3} + C = -\frac{\cos^3 x}{3} + C. \] Also, \[ \int \sin x \, dx = -\cos x + C. \] Therefore, \[ \int \sin^3 x \, dx = -\cos x + \frac{\cos^3 x}{3} + C. \]
Final answer: \[ \boxed{ \int \sin^3 x \, dx = -\cos x + \frac{\cos^3 x}{3} + C. } \]