Q.Evaluate: $\int \dfrac{\sin^{-1}x}{(1-x^2)^{3/2}}\,dx$

The key idea is to use the substitution $x = \sin \theta$, which simplifies both the numerator and the denominator.

Let $x = \sin \theta$, so $dx = \cos \theta \, d\theta$ and $\sin^{-1}x = \theta$. The denominator becomes $(1 - \sin^2 \theta)^{3/2} = (\cos^2 \theta)^{3/2} = \cos^3 \theta$ (taking the positive root since $\theta \in [-\pi/2, \pi/2]$).

The integral transforms as:

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

$\int \theta \sec^2 \theta \, d\theta = \theta \tan \theta - \int \tan \theta \, d\theta = \theta \tan \theta + \log|\cos \theta| + C.$ …

The key insight here is that the denominator $(1-x^2)^{3/2}$ looks like a derivative of something involving $\sin^{-1}x$ or $\cos^{-1}x$. When you see $\sqrt{1-x^2}$ raised to a power, the substitution $x = \sin\theta$ (or $x = \cos\theta$) is almost always the cleanest path — it turns the algebraic mess into a trigonometric playground.

Let’s walk through it.

1. Set up the substitution.

   Let $x = \sin\theta$, so $dx = \cos\theta \, d\theta$.

   The domain: $x \in (-1, 1)$ corresponds to $\theta \in (-\pi/2, \pi/2)$, where $\cos\theta > 0$ — this keeps signs tidy.

2. Rewrite the integrand.

   • $\sin^{-1}x = \theta$ (since $x = \sin\theta$ and $\theta$ is in the principal range).
   • $1 - x^2 = 1 - \sin^2\theta = \cos^2\theta$, so $(1-x^2)^{3/2} = (\cos^2\theta)^{3/2} = |\cos\theta|^3$. Because $\cos\theta > 0$ in our range, $|\cos\theta| = \cos\theta$, so $(1-x^2)^{3/2} = \cos^3\theta$.

   The integral becomes:

3. Integrate by parts.

   This is a classic product: $\theta$ (algebraic) times $\sec^2\theta$ (trigonometric).

   Let $u = \theta$, $dv = \sec^2\theta \, d\theta$.

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

   Integration by parts: $\int u \, dv = uv - \int v \, du$ gives:

4. Handle the leftover integral. $\int \tan\theta \, d\theta = \int \frac{\sin\theta}{\cos\theta} \, d\theta$. Substitute $t = \cos\theta$, $dt = -\sin\theta \, d\theta$, giving:

So:

$\int \theta \sec^2\theta \, d\theta = \theta \tan\theta - (-\log|\cos\theta|) + C = \theta \tan\theta + \log|\cos\theta| + C.$ …

Method: Integration by parts driven by a known derivative

Use this when an integrand is (an inverse function) times (the derivative of a recognisable expression) — for example $\sin^{-1}x$ multiplied by $(1-x^2)^{-3/2}$, which is exactly $\dfrac{d}{dx}\!\left(\dfrac{x}{\sqrt{1-x^2}}\right)$.

Steps

Step 1: Identify the piece that is a clean derivative.

Notice $\dfrac{d}{dx}\!\left(\dfrac{x}{\sqrt{1-x^2}}\right)=\dfrac{1}{(1-x^2)^{3/2}}$. Take this as $dv$, so $v=\dfrac{x}{\sqrt{1-x^2}}$.

Step 2: Set the other factor as $u$ and differentiate.

Step 3: Apply integration by parts $\int u\,dv=uv-\int v\,du$. …

Common Mistakes

Mistake 1: Choosing $u$ and $dv$ badly.

Why it's wrong: taking $u=(1-x^2)^{-3/2}$ leads nowhere; the productive choice is $u=\sin^{-1}x$, $dv=(1-x^2)^{-3/2}dx$ since $v=\tfrac{x}{\sqrt{1-x^2}}$ is clean. Correct approach: let the inverse-trig function be $u$.

Mistake 2: Getting $v$ wrong. …