Integrations

Solution:

Let $I = \int \frac{1}{\sqrt{a^2-x^2}} \, dx$

Substitute $x = a \sin\theta$, which means $\theta = \sin^{-1}\left(\frac{x}{a}\right)$

Differentiating both sides gives: $dx = a \cos\theta \, d\theta$

Substituting these into the integral:

$$ \begin{align*} I &= \int \frac{a \cos\theta}{\sqrt{a^2 – (a \sin\theta)^2}} \, d\theta \\[10pt] I &= \int \frac{a \cos\theta}{\sqrt{a^2 (1 – \sin^2\theta)}} \, d\theta \\[10pt] I &= \int \frac{a \cos\theta}{a \sqrt{\cos^2\theta}} \, d\theta \\[10pt] I &= \int 1 \, d\theta \\[10pt] I &= \theta + c \\[10pt] \text{Therefore, } \int \frac{1}{\sqrt{a^2-x^2}} \, dx &= \sin^{-1}\left(\frac{x}{a}\right) + c \end{align*} $$

---

Hence, evaluating the second part:

We need to find $\int \frac{1}{\sqrt{9-25x^2}} \, dx$. First, factor out $25$ from the square root to match our standard formula form:

$$ \int \frac{1}{\sqrt{25 \left(\frac{9}{25} – x^2\right)}} \, dx = \int \frac{1}{5\sqrt{\left(\frac{3}{5}\right)^2 – x^2}} \, dx $$

Take the constant $\frac{1}{5}$ outside the integral:

$$ = \frac{1}{5} \int \frac{1}{\sqrt{\left(\frac{3}{5}\right)^2 – x^2}} \, dx $$

Using the formula $\int \frac{1}{\sqrt{a^2-x^2}} \, dx = \sin^{-1}\left(\frac{x}{a}\right) + c$, where $a = \frac{3}{5}$:

$$ = \frac{1}{5} \sin^{-1}\left(\frac{x}{\frac{3}{5}}\right) + c = \frac{1}{5} \sin^{-1}\left(\frac{5x}{3}\right) + c $$