Q.Evaluate the definite integral: $\int_{0}^{2/3} \frac{dx}{4+9x^2}$ equals (A) $\frac{\pi}{6}$ (B) $\frac{\pi}{12}$ (C) $\frac{\pi}{24}$ (D) $\frac{\pi}{4}$

Concept: U Substitution — the integrand matches the form $\frac{1}{a^2 + u^2}$, whose antiderivative is $\frac{1}{a} \tan^{-1}\frac{u}{a}$.

Step 1: Write the integral as

Step 2: Let $u = 3x$, so $du = 3\,dx$ and $dx = du/3$. When $x=0$, $u=0$; when $x=2/3$, $u=2$.

Step 3: Use the standard formula $\int \frac{du}{a^2+u^2} = \frac{1}{a}\tan^{-1}\frac{u}{a}$ with $a=2$: …

The key to this problem is spotting that the denominator $4 + 9x^2$ looks like $a^2 + u^2$, the classic form for an inverse tangent integral. The standard formula is:

Here, $4$ is $2^2$, and $9x^2$ is $(3x)^2$. So we have $a = 2$ and $u = 3x$. The substitution $u = 3x$ will cleanly match the formula.

Let’s work through it step by step.

1. Set up the substitution.

   Let $u = 3x$. Then $du = 3\,dx$, so $dx = \frac{du}{3}$.

   The limits change: when $x = 0$, $u = 0$; when $x = \frac{2}{3}$, $u = 3 \cdot \frac{2}{3} = 2$.

2. Rewrite the integral.

   Substitute everything into the integral:

3. Apply the inverse tangent formula. Using $\int \frac{du}{a^2 + u^2} = \frac{1}{a} \tan^{-1}\left(\frac{u}{a}\right)$ with $a = 2$:

4. Evaluate the limits. At $u = 2$: $\tan^{-1}\left(\frac{2}{2}\right) = \tan^{-1}(1) = \frac{\pi}{4}$. At $u = 0$: $\tan^{-1}(0) = 0$. So the result is: $\frac{1}{6} \left( \frac{\pi}{4} - 0 \right) = \frac{\pi}{24}$ …

Method: Reduce $\dfrac{1}{a^2+u^2}$ by a linear substitution to the arctan standard form

When a quadratic denominator is a sum of two squares with a coefficient on $x$, factor it as $a^2+(kx)^2$ and substitute $u=kx$ to reach the memorised inverse-tangent integral.

Steps

Step 1: Write the denominator as a sum of squares.

Recognise $A + Bx^2 = (\sqrt A)^2 + (\sqrt B\,x)^2$, i.e. $a^2+u^2$ with $a=\sqrt A$ and $u=\sqrt B\,x$.

Step 2: Substitute $u=\sqrt B\,x$. …

Common Mistakes

Mistake 1: Ignoring the coefficient of $x^2$ and treating the denominator as $2^2+x^2$.

Why it's wrong: $4+9x^2=2^2+(3x)^2$, so the "$u$" is $3x$, not $x$; missing this scales the answer wrongly. Correct approach: substitute $u=3x$, giving the extra $\tfrac13$ from $dx=\tfrac{du}{3}$.

Mistake 2: Forgetting the $\tfrac1a$ in the arctan formula. …