Q.Integrate the following function: $\frac{1}{\sqrt{9-25x^2}}$

The key idea is that the integrand matches the standard form for the inverse sine derivative: $\frac{d}{dx} \left( \sin^{-1} \frac{x}{a} \right) = \frac{1}{\sqrt{a^2 - x^2}}$.

First, rewrite the denominator to fit $a^2 - u^2$:

Now let $u = 5x$, so $du = 5\,dx$, or $dx = \frac{du}{5}$. The integral becomes:

$\int \frac{1}{\sqrt{9 - u^2}} \cdot \frac{du}{5} = \frac{1}{5} \int \frac{du}{\sqrt{3^2 - u^2}}.$ …

The key here is recognising that the integrand $\frac{1}{\sqrt{9-25x^2}}$ is a perfect match for the derivative of the inverse sine function — but only after a small adjustment. The standard formula is:

The trick is to see that $9 = 3^2$ and $25x^2 = (5x)^2$. So the denominator becomes $\sqrt{3^2 - (5x)^2}$. This suggests we set $u = 5x$, which will make the integral fit the formula exactly — but we must also account for $dx$ in terms of $du$.

Let’s work through it step by step.

1. Rewrite the integrand to reveal the pattern The expression under the square root is $9 - 25x^2$. Write it as $3^2 - (5x)^2$:

2. Choose a substitution Let $u = 5x$. Then $du = 5 \, dx$, so $dx = \frac{du}{5}$. The integral becomes:

3. Apply the inverse sine formula Here $a = 3$ and $u$ is our variable. Using the formula:

So:

4. Substitute back Since $u = 5x$, we get: $\frac{1}{5} \sin^{-1}\left(\frac{5x}{3}\right) + C$ …

Method: Reduce to $\displaystyle\int\frac{dx}{\sqrt{a^2-b^2x^2}}$ (arcsine form)

A radical of the shape $\sqrt{a^2-b^2x^2}$ (a positive constant minus a multiple of $x^2$) integrates to an inverse sine after matching the standard formula.

Steps

Step 1: Identify $a$ and $b$.

For $\sqrt{9-25x^2}$, $a^2=9$ (so $a=3$) and $b^2=25$ (so $b=5$).

Step 2: Apply the standard result.

Step 3: Substitute the values. …

Common Mistakes

Mistake 1: Forgetting the $\dfrac1b$ factor in the arcsine result.

Why it's wrong: $\int\frac{dx}{\sqrt{a^2-b^2x^2}}=\frac1b\sin^{-1}(\frac{bx}{a})$; here $b=5$, so the whole thing is scaled by $\tfrac15$. Dropping it gives an answer five times too large. Correct approach: include the $\tfrac15$, giving $\tfrac15\sin^{-1}(\tfrac{5x}{3})+C$.

Mistake 2: Mismatching $a$ and $b$, e.g. writing $\sin^{-1}(\tfrac{3x}{5})$. …