Use reduction formulas to evaluate the integral. \( \int \sin ^{5}(7 x) d x \)

To evaluate the integral \( \int \sin^{5}(7x) \, dx \), we'll use reduction formulas that relate integrals of powers of sine or cosine. First, we can write: \[ \sin^{n}(x) = \sin^{n-2}(x) \cdot (1 - \cos^{2}(x)). \] For \( n = 5 \): \[ \sin^{5}(7x) = \sin^{3}(7x)(1 - \cos^{2}(7x)). \] So, \[ \int \sin^{5}(7x) \, dx = \int \sin^{3}(7x) \, dx - \int \sin^{3}(7x) \cos^{2}(7x) \, dx. \] Now we'll need another reduction for \( \sin^{3}(7x) \): \[ \sin^{3}(x) = \sin(x)(1 - \cos^{2}(x)). \] This gives us: \[ \int \sin^{3}(7x) \, dx = \int \sin(7x)(1 - \cos^{2}(7x)) \, dx. \] Using the substitution \( u = \cos(7x), \, du = -7 \sin(7x) \, dx\), we obtain: \[ \int \sin^{3}(7x) \, dx = -\frac{1}{7} \left( \int (1 - u^{2}) \, du \right). \] Now integrating gives: \[ -\frac{1}{7} \left( u - \frac{u^{3}}{3} \right) = -\frac{1}{7} \left( \cos(7x) - \frac{\cos^{3}(7x)}{3} \right). \] Therefore, combining all results, we can derive: \[ \int \sin^{5}(7x) \, dx = \frac{1}{7} \left( \cos(7x) - \frac{\cos^{3}(7x)}{3} \right) + C, \] where \( C \) is the constant of integration.