Q.Integrate the following function: $\frac{\sin^{-1}x}{\sqrt{1 - x^2}}$

Concept: U Substitution — the derivative of $\sin^{-1}x$ appears in the denominator, making it a natural choice for substitution.

Let $u = \sin^{-1}x$. Then $du = \frac{1}{\sqrt{1 - x^2}}\,dx$.

The integral becomes:

$\int u \, du = \frac{u^2}{2} + C$ …

Why U-Substitution Works Here

When you see a composition like $\sin^{-1} x$ inside the numerator and its derivative $\frac{1}{\sqrt{1 - x^2}}$ lurking in the denominator, that’s a flashing neon sign for substitution. The derivative of $\sin^{-1} x$ is exactly $\frac{1}{\sqrt{1 - x^2}}$, so setting $u = \sin^{-1} x$ will turn the whole mess into something clean.

Let’s walk through it.

1. Choose the substitution. Let $u = \sin^{-1} x$. Then differentiate:

This means $du = \frac{dx}{\sqrt{1 - x^2}}$.

2. Rewrite the integral. The original integral is

Replace $\sin^{-1} x$ with $u$, and $\frac{dx}{\sqrt{1 - x^2}}$ with $du$:

3. Integrate. This is a basic power rule:

4. Substitute back. Recall $u = \sin^{-1} x$, so: $\frac{(\sin^{-1} x)^2}{2} + C$ …

Method: Substitute an inverse-trig function paired with its derivative

Use this when $\sin^{-1}x$ (or another inverse-trig function) appears together with $\dfrac{1}{\sqrt{1-x^2}}$, its own derivative.

Steps

Step 1: Recognise the derivative pairing.

$\dfrac{d}{dx}\big(\sin^{-1}x\big)=\dfrac{1}{\sqrt{1-x^2}}$, which multiplies $\sin^{-1}x$ in the integrand.

Step 2: Substitute $u=\sin^{-1}x$.

Then $du=\dfrac{1}{\sqrt{1-x^2}}\,dx$, reducing the integral to $\displaystyle\int u\,du$. …

Common Mistakes

Mistake 1: Not recognising $\dfrac{1}{\sqrt{1-x^2}}$ as the derivative of $\sin^{-1}x$.

Why it's wrong: missing this hides the substitution $u=\sin^{-1}x$. Correct approach: recall $\dfrac{d}{dx}\sin^{-1}x=\dfrac{1}{\sqrt{1-x^2}}$.

Mistake 2: Trying to integrate $\sin^{-1}x$ and $\dfrac{1}{\sqrt{1-x^2}}$ separately.

Why it's wrong: they form one $u\,du$ package. Correct approach: substitute so the integral becomes $\int u\,du$. …