Q.Evaluate: $\int_{1}^{2} \dfrac{dx}{\sqrt{(x-1)(2-x)}}$

The key idea is to rewrite the integrand so it matches the form $\frac{1}{\sqrt{1 - u^2}}$, which integrates to $\sin^{-1} u$.

First, complete the square inside the square root:

So the integral becomes

Now use the substitution $u = \frac{x - 3/2}{1/2} = 2x - 3$, so $dx = \frac{1}{2} du$. When $x=1$, $u=-1$; when $x=2$, $u=1$. The integral transforms to …

We start with the integral

The expression under the square root is a product of two linear factors that vanish at the endpoints. That’s a strong hint: the integrand blows up at both limits, but the integral converges. The standard trick is to rewrite the product as a difference of squares.

1. Rewrite the product Expand $(x-1)(2-x)$:

Factor out the negative sign:

Complete the square for $x^2 - 3x$:

So

Therefore

2. Recognise the standard form The integrand becomes

This matches $\frac{1}{\sqrt{a^2 - u^2}}$ with $a = \frac{1}{2}$ and $u = x - \frac{3}{2}$. The antiderivative is $\sin^{-1}\left(\frac{u}{a}\right)$.

3. Substitute and integrate Let $u = x - \frac{3}{2}$, so $du = dx$. When $x = 1$, $u = -\frac{1}{2}$; when $x = 2$, $u = \frac{1}{2}$.

The integral of $\frac{1}{\sqrt{a^2 - u^2}}$ is $\sin^{-1}\left(\frac{u}{a}\right)$. Here $a = \frac{1}{2}$, so …

Method: Completing the square for $\int\frac{dx}{\sqrt{(x-\alpha)(\beta-x)}}$

Use this for definite integrals whose denominator is the root of a product like $(x-1)(2-x)$. Expanding and completing the square gives an arcsine form with a clean $\pi$-type value.

Steps

Step 1: Expand the product.

Step 2: Complete the square.

So $a=\tfrac12$, $u=x-\tfrac32$.

Step 3: Apply the arcsine standard form.

Step 4: Evaluate at the limits. …

Common Mistakes

Mistake 1: Not expanding before completing the square.

Why it's wrong: $(x-1)(2-x)$ must be expanded to $-x^2+3x-2$ before the square can be completed. Correct approach: multiply out first.

Mistake 2: Sign error giving $u^2-a^2$.

Why it's wrong: the correct form is $\tfrac14-(x-\tfrac32)^2$ (an arcsine), not $(x-\tfrac32)^2-\tfrac14$. Correct approach: factor the $-1$ from the $x^2$ term carefully.

Mistake 3: Evaluating $\sin^{-1}(-1)$ wrongly. …