Q.Integrate the following function: $\int \frac{dx}{\sqrt{9x - 4x^2}}$ equals (A) $\frac{1}{9} \sin^{-1} \left(\frac{9x - 8}{8}\right) + C$ (B) $\frac{1}{2} \sin^{-1} \left(\frac{8x - 9}{9}\right) + C$ (C) $\frac{1}{3} \sin^{-1} \left(\frac{9x - 8}{8}\right) + C$ (D) $\frac{1}{2} \sin^{-1} \left(\frac{9x - 8}{9}\right) + C$

Concept: Integration by completing the square — rewrite the quadratic inside the square root in the form $a^2 - (x - b)^2$ to use the standard integral $\int \frac{dx}{\sqrt{a^2 - u^2}} = \sin^{-1}\frac{u}{a} + C$.

Step 1: Factor the quadratic.

$9x - 4x^2 = -4\left(x^2 - \frac{9}{4}x\right) = -4\left[\left(x - \frac{9}{8}\right)^2 - \frac{81}{64}\right] = 4\left[\frac{81}{64} - \left(x - \frac{9}{8}\right)^2\right]$.

Step 2: Take the constant out of the square root.

$\sqrt{9x - 4x^2} = 2\sqrt{\frac{81}{64} - \left(x - \frac{9}{8}\right)^2}$.

Step 3: Integrate using the standard form. …

The key here is that the expression under the square root, $9x - 4x^2$, is a quadratic that does not fit a standard integration formula directly. But if we rewrite it as a perfect square minus something, we can turn it into the form $a^2 - u^2$, which integrates to an inverse sine.

Why does completing the square work? Because the derivative of $\sin^{-1}(x)$ involves $\frac{1}{\sqrt{1-x^2}}$, and any quadratic under a square root can be manipulated into that shape by shifting and scaling the variable. The constant $a$ becomes the "radius" of the sine-arc.

Let's go step by step.

1. Factor out the coefficient of $x^2$ to make the square easier to complete.

The quadratic is $-4x^2 + 9x$. Factor $-4$ from the first two terms:

We'll complete the square inside the parentheses.

2. Complete the square for $x^2 - \frac{9}{4}x$.

Take half of $-\frac{9}{4}$: that's $-\frac{9}{8}$. Square it: $\frac{81}{64}$. Add and subtract this inside:

3. Substitute back into the original expression.

So the quadratic becomes:

4. Factor out $\frac{81}{16}$ to reveal the $1 - u^2$ form.

Write:

Simplify the coefficient: $\frac{64}{81} = \left(\frac{8}{9}\right)^2$. So:

5. Take the square root (positive, since it's a length in the integral).

Because $\frac{8}{9}\left(x - \frac{9}{8}\right) = \frac{8}{9}x - 1$.

6. Substitute into the integral. …

Method: Integral of $\dfrac{1}{\sqrt{\text{quadratic}}}$ with a negative $x^2$ term (arcsine form)

Use this for $\displaystyle\int\frac{dx}{\sqrt{\alpha x-\beta x^2}}$ or any $\dfrac{1}{\sqrt{\text{quadratic}}}$ whose $x^2$-coefficient is negative — the target is the standard $\sin^{-1}$ integral.

Steps

Step 1: Factor out the (negative) leading coefficient so the bracket has a $+1$ on $x^2$:

Step 2: Complete the square inside and simplify the signs to reach an $a^2-u^2$ shape:

$= \beta\left[k^2-(x-h)^2\right].$ …

Common Mistakes

Mistake 1: Completing the square without first handling the $-4$ coefficient of $x^2$.

Why it's wrong: $9x-4x^2$ has leading coefficient $-4$; completing the square as if it were $9x-x^2$ misplaces both the centre and the "radius", giving wrong constants (distractors A and C with $\tfrac19$ or $\tfrac13$ out front). Correct approach: factor out $-4$ first, then complete the square inside.

Mistake 2: Losing the $\tfrac12$ that comes from $\sqrt4=2$. …