91Ó°ÊÓ

We should first identify the type of integral we are dealing with. In this case, we have a rational function containing a square root in the denominator, which is an inverse trigonometric integral. To be more specific, this integral is an arcsecant type.

To solve this type of integral, we will use a trigonometric substitution. For arcsecant type with square root of the form \(\sqrt{a^2 - x^2}\), we should use the substitution formula: \(x = a\cdot \sin{\theta}\). In our case, \(a = 5\). Thus, the substitution is \(x = 5\sin{\theta}\). We also need to find the differential \(d x\), which is \(d x = 5\cos{\theta} d\theta\).

Now we will substitute \(x\) and \(d x\) into the integral: $$\int \frac{6}{\sqrt{25-x^{2}}} d x = \int \frac{6}{\sqrt{25 - (5\sin{\theta})^2}}\cdot 5\cos{\theta} d\theta$$

Simplify the integral by cancelling terms and evaluating the expression under the square root: $$\int \frac{6}{\sqrt{25 - (5\sin{\theta})^2}}\cdot 5\cos{\theta} d\theta = \int \frac{6\cdot 5\cos{\theta}}{\sqrt{25 - 25\sin^2{\theta}}} d\theta = \int \frac{30\cos{\theta}}{\sqrt{25(1 - \sin^2{\theta})}} d\theta$$ We know that \(\cos^2{\theta} = 1 - \sin^2{\theta}\). Therefore, we can rewrite the expression: $$\int \frac{30\cos{\theta}}{\sqrt{25(1 - \sin^2{\theta})}} d\theta = \int \frac{30\cos{\theta}}{\sqrt{25\cos^2{\theta}}} d\theta$$ Now, we can further simplify the integral by cancelling \(25\cos^2{\theta}\) in the denominator with the remaining terms in the numerator: $$\int \frac{30\cos{\theta}}{\sqrt{25\cos^2{\theta}}} d\theta = \int \frac{30\cos{\theta}}{5\cos{\theta}} d\theta = 6\int d\theta$$

Evaluating the integral, we get: $$6\int d\theta = 6\theta + C$$ where C is the constant of integration.

Since our original variable was \(x\), we need to convert \(\theta\) back into terms of \(x\). From our initial substitution, \(x = 5\sin{\theta}\), we have \(\sin{\theta} = \frac{x}{5}\). To find \(\theta\), we can use the inverse sine function: $$\theta = \arcsin{\left(\frac{x}{5}\right)}$$ Substituting this back into our solution, we get: $$6\theta + C = 6\arcsin{\left(\frac{x}{5}\right)} + C$$ This is the indefinite integral of the given function.

To verify the result, we will differentiate our solution and check whether it returns the original function: $$\frac{d}{dx}\left(6\arcsin{\left(\frac{x}{5}\right)} + C\right) = \frac{6}{\sqrt{1 - \left(\frac{x}{5}\right)^2}} \cdot \frac{1}{5} = \frac{6}{\sqrt{25 - x^2}}$$ Since the derivative returns the original function, the integral solution is correct.