Q.Integrate the function $\frac{1}{x\sqrt{ax-x^2}}\ \text{[Hint: Put } x=\frac{a}{t}\text{]}$

Concept: U Substitution — the hint $x = \frac{a}{t}$ simplifies the square root.

Step 1: Substitute $x = \frac{a}{t}$, so $dx = -\frac{a}{t^2} dt$. Also, $ax - x^2 = a\left(\frac{a}{t}\right) - \frac{a^2}{t^2} = \frac{a^2}{t} - \frac{a^2}{t^2} = \frac{a^2(t-1)}{t^2}$.

Step 2: The integrand becomes:

Simplify: $\frac{1}{\frac{a}{t} \cdot \frac{a\sqrt{t-1}}{t}} = \frac{t^2}{a^2 \sqrt{t-1}}$. Multiply by $-\frac{a}{t^2} dt$ gives $-\frac{1}{a\sqrt{t-1}} dt$.

Step 3: Integrate:

Step 4: Substitute back $t = \frac{a}{x}$:

Why This Substitution Works

When you see $\sqrt{ax - x^2}$, your first instinct might be to complete the square: $ax - x^2 = \frac{a^2}{4} - \left(x - \frac{a}{2}\right)^2$. That’s a valid path, but it leads to a trigonometric substitution. The hint suggests a different, cleverer route: put $x = \frac{a}{t}$. Why?

Notice that $ax - x^2 = x(a - x)$. If we set $x = a/t$, then $a - x = a - a/t = a(1 - 1/t) = a \cdot \frac{t-1}{t}$. The product becomes:

The square root then gives $\sqrt{ax - x^2} = \frac{a}{t} \sqrt{t-1}$, and the $x$ in the denominator outside the root cancels beautifully. The substitution turns a complicated radical into something you can integrate with a simple power rule.

Step-by-Step Solution

1. Set up the substitution.

Let $x = \frac{a}{t}$, where $a$ is a constant (presumably $a > 0$ for the square root to be real). Then differentiate:

2. Rewrite the integrand in terms of $t$.

The integrand is $\frac{1}{x \sqrt{ax - x^2}}$. First, $x$ in the denominator becomes $a/t$. Next, the expression under the square root:

So,

taking the positive root (we assume $t > 1$ or $t < 0$ as needed for the domain).

3. Combine everything.

The integrand becomes:

Now include $dx = -\frac{a}{t^2} dt$:

4. Integrate with respect to $t$.

The integral is now straightforward:

Using the power rule, $\int (t-1)^{-1/2} \, dt = 2 (t-1)^{1/2} + C$. So,

5. Substitute back to $x$.

Recall $x = a/t$, so $t = a/x$. Then $t-1 = \frac{a}{x} - 1 = \frac{a - x}{x}$. Therefore,

The final antiderivative is:

Method: Reciprocal substitution $x=\tfrac{a}{t}$ for $\int\frac{dx}{x\sqrt{ax-x^2}}$

Use this for an integrand with a lone $x$ (or $x^2$) multiplying a square root of a quadratic — replacing $x$ by $a/t$ makes the outside factor cancel the transformed radical.

Steps

Step 1: Set the substitution and its differential.

Let $x=\dfrac{a}{t}$, so $dx=-\dfrac{a}{t^2}\,dt$. Carry the minus sign — dropping it is the classic error here.

Step 2: Rewrite the radical.

Factor the quadratic as $ax-x^2=x(a-x)$ and substitute; the square root simplifies to a single power of $t$ times a constant, and the extra powers of $x$ in the integrand cancel.

Step 3: Integrate the reduced form and back-substitute.

You reach a standard power integral in $t$ (typically $\int (t-1)^{-1/2}\,dt=2\sqrt{t-1}$). Finish by replacing $t=\dfrac{a}{x}$ to return to $x$.

Common Mistakes

Mistake 1: Dropping the minus sign in $dx=-\tfrac{a}{t^2}\,dt$.

Why it's wrong: the whole final sign hinges on it; losing it gives $+\tfrac{2}{a}\sqrt{\cdots}$ instead of the correct negative. Correct approach: substitute $dx$ with its minus sign explicitly.

Mistake 2: Mishandling $\sqrt{\dfrac{a^2(t-1)}{t^2}}$.

Why it's wrong: it equals $\dfrac{a}{t}\sqrt{t-1}$ (for the relevant domain); a botched simplification leaves stray $t$'s that don't cancel. Correct approach: simplify the radical carefully and confirm the $t^2$ factors cancel.

Mistake 3: Forgetting to return to $x$.

Why it's wrong: the answer must be in $x$; leaving $\sqrt{t-1}$ is incomplete. Correct approach: use $t=\tfrac{a}{x}$ so $\sqrt{t-1}=\sqrt{\tfrac{a-x}{x}}$.