use a change of variables to find the following Indefinite intergal \( \int \frac{x}{\sqrt{1-25 x^{2}}} d x= \)

To tackle the integral \( \int \frac{x}{\sqrt{1-25 x^{2}}} dx \), we can use a trigonometric substitution. Let's set \( x = \frac{1}{5} \sin(\theta) \). This implies that \( dx = \frac{1}{5} \cos(\theta) d\theta \), and the integral transforms as follows: The expression under the square root becomes: \[ 1 - 25x^2 = 1 - 25\left(\frac{1}{5} \sin(\theta)\right)^2 = 1 - \sin^2(\theta) = \cos^2(\theta). \] Thus, we have: \[ \sqrt{1-25 x^{2}} = \cos(\theta). \] Now substituting everything into the integral gives: \[ \int \frac{\frac{1}{5} \sin(\theta)}{\cos(\theta)} \cdot \frac{1}{5} \cos(\theta) d\theta = \int \frac{1}{25} \sin(\theta) d\theta. \] Integrating \( \int \frac{1}{25} \sin(\theta) d\theta \) gives: \[ -\frac{1}{25} \cos(\theta) + C. \] Finally, we need to revert back to the variable \( x \). Recall that we set \( x = \frac{1}{5} \sin(\theta) \), which means \( \sin(\theta) = 5x \). Thus, \( \cos(\theta) = \sqrt{1 - \sin^2(\theta)} = \sqrt{1 - (5x)^2} = \sqrt{1 - 25x^2} \). Putting it all together, we have: \[ \int \frac{x}{\sqrt{1-25 x^{2}}} dx = -\frac{1}{25} \sqrt{1 - 25x^2} + C. \]