Q.Evaluate: $\int \dfrac{dx}{\sqrt{16-9x^2}}$

Concept: Integration by completing the square (here, the square is already complete — this is a standard $\int \frac{dx}{\sqrt{a^2 - x^2}}$ form after factoring).

Step 1: Rewrite the denominator to match the form $\sqrt{a^2 - u^2}$.

Step 2: Substitute $u = x$, so $dx = du$, and factor out the constant from the square root.

$\int \frac{dx}{\sqrt{16-9x^2}} = \int \frac{dx}{\sqrt{9\left(\frac{16}{9} - x^2\right)}} = \frac{1}{3} \int \frac{dx}{\sqrt{\left(\frac{4}{3}\right)^2 - x^2}}$ …

The key to integrals like $\int \frac{dx}{\sqrt{16-9x^2}}$ is recognising the pattern $\sqrt{a^2 - u^2}$. When you see a quadratic under a square root, your first instinct should be: can I write it as $a^2 - (\text{something})^2$? If yes, the integral becomes an inverse sine.

Here, $16-9x^2$ is already a difference of squares: $16 = 4^2$ and $9x^2 = (3x)^2$. So we have $\sqrt{4^2 - (3x)^2}$. That’s exactly the form $\sqrt{a^2 - u^2}$ with $a = 4$ and $u = 3x$.

But there’s a catch: the $dx$ is in terms of $x$, while the formula expects $du$. So we need a substitution.

1. Set up the substitution. Let $u = 3x$. Then $du = 3\,dx$, so $dx = \frac{du}{3}$. The integral becomes:

2. Apply the standard formula. The formula $\int \frac{du}{\sqrt{a^2 - u^2}} = \sin^{-1}\frac{u}{a} + C$ is a direct consequence of the derivative $\frac{d}{du}\sin^{-1}\frac{u}{a} = \frac{1}{\sqrt{a^2 - u^2}}$. Here $a = 4$, so:

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

Method: Reducing to the standard form $\int\frac{dx}{\sqrt{a^2-u^2}}=\sin^{-1}\frac{u}{a}$

Use this whenever a denominator is $\sqrt{(\text{constant})-(\text{coefficient})x^2}$. Factor out the coefficient so it matches the arcsine standard form.

Steps

Step 1: Factor the coefficient of $x^2$ from under the root.

Step 2: Identify $a$.

Here $a=\tfrac43$, so the integral becomes $\tfrac13\displaystyle\int\frac{dx}{\sqrt{a^2-x^2}}$.

Step 3: Apply the standard result. …

Common Mistakes

Mistake 1: Not factoring out the $9$.

Why it's wrong: applying arcsine to $\sqrt{16-9x^2}$ without extracting the coefficient gives a wrong $a$ and a missing $\tfrac13$. Correct approach: write $16-9x^2=9(\tfrac{16}{9}-x^2)$ first.

Mistake 2: Wrong value of $a$.

Why it's wrong: after factoring, $a^2=\tfrac{16}{9}$ so $a=\tfrac43$; using $a=4$ or $a=\tfrac34$ gives the wrong argument. Correct approach: take $a=\sqrt{16/9}=\tfrac43$, so the argument is $\tfrac{3x}{4}$. …