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

To evaluate the integral \( \int \sin^5(5x) \, dx \), we can use a reduction formula for sine integrals. The reduction formula for \( \int \sin^n(x) \, dx \) is given by: \[ \int \sin^n(x) \, dx = -\frac{\sin^{n-1}(x)\cos(x)}{n} + \frac{n-1}{n} \int \sin^{n-2}(x) \, dx \] Let's adapt this to our integral: 1. Set \( n = 5 \) and factor out the constant multiplier from the angle: \[ \int \sin^5(5x) \, dx = \frac{1}{5} \int \sin^5(u) \, du \quad \text{where } u = 5x \text{ and } du = 5dx \] This gives us: \[ = \frac{1}{5} \int \sin^5(u) \, du \] 2. Applying the reduction formula: \[ \int \sin^5(u) \, du = -\frac{\sin^4(u) \cos(u)}{5} + \frac{4}{5} \int \sin^3(u) \, du \] 3. Next, we need \( \int \sin^3(u) \, du \) which can also be evaluated using the reduction formula: \[ \int \sin^3(u) \, du = -\frac{\sin^2(u) \cos(u)}{3} + \frac{2}{3} \int \sin(u) \, du \] 4. The integral \( \int \sin(u) \, du \) is straightforward: \[ \int \sin(u) \, du = -\cos(u) \] 5. Putting it all together: Substitute back into the equation for \( \int \sin^3(u) \, du \): \[ \int \sin^3(u) \, du = -\frac{\sin^2(u) \cos(u)}{3} - \frac{2}{3} \cos(u) \] 6. Now substitute \( \int \sin^3(u) \) into the \( \int \sin^5(u) \): \[ \int \sin^5(u) \, du = -\frac{\sin^4(u) \cos(u)}{5} + \frac{4}{5} \left(-\frac{\sin^2(u) \cos(u)}{3} - \frac{2}{3} \cos(u)\right) \] 7. Now we combine all parts: \[ \int \sin^5(5x) \, dx = \frac{1}{5} \left(-\frac{\sin^4(5x) \cos(5x)}{5} + \frac{4}{5} \left(-\frac{\sin^2(5x) \cos(5x)}{3} - \frac{2}{3} \cos(5x)\right)\right) + C \] This yields the final result involving multiple hyperbolic trigonometric identities and integration by parts. Finally, simplify and remember to substitute back in \( x \) when you finish the integration! And voilà! You've evaluated the integral \( \int \sin^5(5x) \, dx \)!