Question

Evaluate $\int (\sqrt{25-9x^2})dx$.

Answer 1

$\frac{x}{2}\sqrt{25 - 9x^2} + \frac{25}{6}\sin^{-1}\left(\frac{3x}{5}\right) + C$

Answer 2

The integral to evaluate is $\int (\sqrt{25-9x^2})dx$.

Explanation of the solution:

1. Rewrite the integrand: The expression inside the square root can be written as $5^2 - (3x)^2$. So, the integral is $\int \sqrt{5^2 - (3x)^2} dx$.

2. Substitution: Let $u = 3x$. Differentiating both sides with respect to $x$, we get $du = 3 dx$. This implies $dx = \frac{1}{3} du$.

3. Transform the integral: Substitute $u$ and $dx$ into the integral: $\int \sqrt{5^2 - u^2} \left(\frac{1}{3} du\right) = \frac{1}{3} \int \sqrt{5^2 - u^2} du$

4. Apply standard integral formula: The integral is now in the standard form $\int \sqrt{a^2 - x^2} dx$, where $a=5$ and the variable is $u$. The formula for this integral is: $\int \sqrt{a^2 - x^2} dx = \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}\right) + C$ Applying this formula with $a=5$ and variable $u$: $\frac{1}{3} \left[ \frac{u}{2}\sqrt{5^2 - u^2} + \frac{5^2}{2}\sin^{-1}\left(\frac{u}{5}\right) \right] + C$

5. Substitute back: Replace $u$ with $3x$: $\frac{1}{3} \left[ \frac{3x}{2}\sqrt{25 - (3x)^2} + \frac{25}{2}\sin^{-1}\left(\frac{3x}{5}\right) \right] + C$

6. Simplify: $\frac{1}{3} \left[ \frac{3x}{2}\sqrt{25 - 9x^2} + \frac{25}{2}\sin^{-1}\left(\frac{3x}{5}\right) \right] + C$ $= \frac{3x}{6}\sqrt{25 - 9x^2} + \frac{25}{6}\sin^{-1}\left(\frac{3x}{5}\right) + C$ $= \frac{x}{2}\sqrt{25 - 9x^2} + \frac{25}{6}\sin^{-1}\left(\frac{3x}{5}\right) + C$