Q.Integrate the following function: $\frac{1}{\sqrt{(x-a)(x-b)}}$

The key idea is to rewrite the product under the square root into a difference of squares, then use a trigonometric substitution.

First, complete the square for the quadratic under the root:

Let $u = x - \frac{a+b}{2}$, so $du = dx$ and the integral becomes:

This is a standard form. Using the substitution $u = \frac{a-b}{2} \sec \theta$ (or directly recalling the formula), we get: …

Let’s start with the concept. You’re asked to integrate $\frac{1}{\sqrt{(x-a)(x-b)}}$. At first glance, the product inside the square root looks like a quadratic in $x$: $(x-a)(x-b) = x^2 - (a+b)x + ab$. This is a quadratic expression, and integrals of the form $\frac{1}{\sqrt{\text{quadratic}}}$ are classic candidates for a substitution that turns them into a standard form like $\int \frac{du}{\sqrt{u^2 \pm k^2}}$, which integrates to an inverse hyperbolic sine (or a logarithm). The trick is to complete the square to reveal a perfect square plus or minus a constant.

Why does this work? Because the derivative of the expression inside the square root often appears in the numerator after a clever substitution, or we can use a trigonometric/hyperbolic substitution. Here, the most efficient path is to shift the variable to center the quadratic, then use a substitution that simplifies the square root into something like $\sqrt{t^2 - d^2}$.

Let’s work through it step by step.

1. Rewrite the integrand by expanding and completing the square.

   $(x-a)(x-b) = x^2 - (a+b)x + ab$.

   Complete the square:

   $x^2 - (a+b)x + ab = \left(x - \frac{a+b}{2}\right)^2 - \left(\frac{a+b}{2}\right)^2 + ab$.

   Simplify the constant term:

   $-\left(\frac{a+b}{2}\right)^2 + ab = -\frac{a^2 + 2ab + b^2}{4} + ab = \frac{-a^2 - 2ab - b^2 + 4ab}{4} = \frac{-a^2 + 2ab - b^2}{4} = -\frac{(a-b)^2}{4}$.

   So $(x-a)(x-b) = \left(x - \frac{a+b}{2}\right)^2 - \left(\frac{a-b}{2}\right)^2$.

2. Make a substitution to center the variable.

   Let $t = x - \frac{a+b}{2}$. Then $dt = dx$, and the integral becomes

   $\int \frac{dt}{\sqrt{t^2 - \left(\frac{a-b}{2}\right)^2}}$.

   Notice the constant $d = \frac{|a-b|}{2}$ (we’ll assume $a \neq b$; if $a=b$, the integrand is $\frac{1}{|x-a|}$, which integrates to $\log|x-a| + C$). For definiteness, let $c = \frac{a-b}{2}$, so the integral is $\int \frac{dt}{\sqrt{t^2 - c^2}}$.

3. Recognize the standard form.

   The integral $\int \frac{dt}{\sqrt{t^2 - c^2}}$ is a standard result. You can derive it using a hyperbolic substitution: let $t = c \cosh u$, then $dt = c \sinh u \, du$, and $\sqrt{t^2 - c^2} = c \sinh u$ (for $t > c$). The integral becomes $\int \frac{c \sinh u \, du}{c \sinh u} = \int du = u + C = \cosh^{-1}\left(\frac{t}{c}\right) + C$.

   Alternatively, the result is often written as $\log\left| t + \sqrt{t^2 - c^2} \right| + C$, which is valid for $|t| > |c|$.

   $\int \frac{dt}{\sqrt{t^2 - c^2}} = \log\left| t + \sqrt{t^2 - c^2} \right| + C$

4. Substitute back in terms of $x$.

   Recall $t = x - \frac{a+b}{2}$ and $c = \frac{a-b}{2}$. Then …

Method: $1/\sqrt{(x-a)(x-b)}$ — complete the square with literal constants

A product of two linear factors under the root is still a quadratic; the same complete-the-square method works even when the roots are letters, not numbers.

Steps

Step 1: Expand the product.

Step 2: Complete the square. The vertex shift is half the linear coefficient:

$(x-a)(x-b) = \left(x-\tfrac{a+b}{2}\right)^2 - \left(\tfrac{a-b}{2}\right)^2 .$ …

Common Mistakes

Mistake 1: Mis-computing the constant term after completing the square.

Why it's wrong: $ab - \left(\tfrac{a+b}{2}\right)^2 = -\left(\tfrac{a-b}{2}\right)^2$, not $-\left(\tfrac{a+b}{2}\right)^2$. Correct approach: simplify $ab-\tfrac{(a+b)^2}{4}$ fully to $-\tfrac{(a-b)^2}{4}$.

Mistake 2: Choosing the $\sin^{-1}$ form.

Why it's wrong: the completed form is $u^2 - k^2$ (positive $x^2$), so the answer is a logarithm, not an inverse sine. Correct approach: reserve $\sin^{-1}$ for $k^2-u^2$. …