Q.$\int \sin^{-1}\sqrt{\dfrac{x}{a+x}}\,dx$ (Hint: Put $x=a\tan^2\theta$)

Concept: Power Rule Integration — after substitution, the integrand simplifies to a standard trigonometric form.

Let $x = a \tan^2 \theta$, so $dx = 2a \tan \theta \sec^2 \theta \, d\theta$.

Then $\sqrt{\frac{x}{a+x}} = \sqrt{\frac{a\tan^2\theta}{a + a\tan^2\theta}} = \sqrt{\frac{\tan^2\theta}{\sec^2\theta}} = \sin\theta$ (taking $\theta \in [0, \pi/2)$).

Thus $\sin^{-1}(\sin\theta) = \theta$, and the integral becomes:

Now integrate by parts: let $u = \theta$, $dv = \tan\theta \sec^2\theta \, d\theta$.

Then $du = d\theta$, and $v = \frac{1}{2}\tan^2\theta$ (since $\frac{d}{d\theta}(\tan^2\theta) = 2\tan\theta\sec^2\theta$).

So:

$2a \left[ \frac{\theta}{2}\tan^2\theta - \frac{1}{2}\int \tan^2\theta \, d\theta \right] = a\theta\tan^2\theta - a\int (\sec^2\theta - 1)\, d\theta.$ …

The problem asks for the indefinite integral of $\sin^{-1}\sqrt{\frac{x}{a+x}}$. The expression inside the inverse sine looks messy, but the hint suggests a clever substitution: $x = a \tan^2 \theta$. Why does this work? Because $\sqrt{\frac{x}{a+x}}$ becomes something like $\sqrt{\frac{a \tan^2 \theta}{a + a \tan^2 \theta}} = \sqrt{\frac{\tan^2 \theta}{1 + \tan^2 \theta}} = \sqrt{\sin^2 \theta} = \sin \theta$ (assuming $\theta$ in a suitable range). Then $\sin^{-1}(\sin \theta) = \theta$, which is much simpler to integrate.

Let’s walk through it step by step.

1. Substitute $x = a \tan^2 \theta$. We need $dx$ in terms of $d\theta$. Differentiate:

Also, note that $a + x = a + a \tan^2 \theta = a(1 + \tan^2 \theta) = a \sec^2 \theta$.

2. Simplify the integrand. Compute $\sqrt{\frac{x}{a+x}}$:

taking $\theta \in [0, \pi/2)$ so $\sin \theta \ge 0$.

Hence $\sin^{-1}\sqrt{\frac{x}{a+x}} = \sin^{-1}(\sin \theta) = \theta$.

3. Rewrite the integral. The integral becomes:

4. Simplify the trigonometric part. Notice $\tan \theta \sec^2 \theta = \frac{\sin \theta}{\cos^3 \theta}$. But a better approach: let $t = \tan \theta$, then $dt = \sec^2 \theta \, d\theta$, so $\tan \theta \sec^2 \theta \, d\theta = t \, dt$. However, we still have $\theta$ in terms of $t$: $\theta = \tan^{-1} t$. So:

5. Integrate by parts. Let $u = \tan^{-1} t$ and $dv = t \, dt$. Then $du = \frac{1}{1+t^2} dt$ and $v = \frac{t^2}{2}$. Integration by parts gives:

so:

6. Simplify the remaining integral. Write $\frac{t^2}{1+t^2} = 1 - \frac{1}{1+t^2}$. Then:

Substitute back: …

Method: Rationalising a nested radical by trig substitution

When an inverse-trig function wraps a radical like $\sqrt{\tfrac{x}{a+x}}$, a substitution $x=a\tan^2\theta$ collapses the radical to a plain sine; you then integrate by parts.

Steps

Step 1: Substitute to simplify the inner function.

With $x=a\tan^2\theta$, $a+x=a\sec^2\theta$, so $\sqrt{\tfrac{x}{a+x}}=\sin\theta$ and $\sin^{-1}(\sin\theta)=\theta$. Compute $dx=2a\tan\theta\sec^2\theta\,d\theta$.

Step 2: Integrate by parts. …

Common Mistakes

Mistake 1: Simplifying $\sqrt{\tfrac{x}{a+x}}$ to the wrong trig function.

Why it's wrong: with $x=a\tan^2\theta$, $\frac{x}{a+x}=\frac{\tan^2\theta}{\sec^2\theta}=\sin^2\theta$, so the root is $\sin\theta$ — not $\tan\theta$ or $\cos\theta$. Correct approach: simplify $a+x=a\sec^2\theta$ first.

Mistake 2: Losing the $2a\tan\theta\sec^2\theta$ factor from $dx$. …