To evaluate \( \int \sin^5 x \, dx \), we rewrite the integral as: \[ \int \sin^5 x \, dx = \int \sin^3 x \cdot \sin^2 x \, dx = \int \sin^3 x \cdot (1 - \cos^2 x) \, dx \] Substituting \( \sin^3 x = \sin x (1 - \cos^2 x) \): \[ = \int \sin x (1 - \cos^2 x)^2 (1 - \cos^2 x) \, dx = \int \sin x (1 - 2 \cos^2 x + \cos^4 x) \, dx \] Let \( u = \cos x \). Then \( du = -\sin x \, dx \). The integral transforms to: \[ = -\int (1 - 2u^2 + u^4) \, du = -\left( u - \frac{2u^3}{3} + \frac{u^5}{5} \right) + C \] Substituting back \( u = \cos x \): \[ = -\cos x + \frac{2}{3} \cos^3 x - \frac{1}{5} \cos^5 x + C \]