Q.Evaluate the definite integral: $\int_0^1 \frac{dx}{\sqrt{1-x^2}}$

The key idea is that the integrand $\frac{1}{\sqrt{1-x^2}}$ is the derivative of $\sin^{-1} x$, so we can directly apply the standard formula.

Step 1: Recall the antiderivative:

Step 2: Evaluate the definite integral from $0$ to $1$: …

Why This Integral Works

The expression $\frac{1}{\sqrt{1-x^2}}$ is a classic derivative — it’s the derivative of $\sin^{-1} x$ (or $\arcsin x$). When you see an integrand that matches a known derivative form, the integration becomes immediate: you’re just reversing the differentiation. The limits $0$ to $1$ are especially nice because $\sin^{-1} x$ is defined on $[-1, 1]$, and at $x=1$, it hits $\frac{\pi}{2}$.

The key insight: recognise the derivative pattern. No substitution is needed here — it’s a direct antiderivative. But if you wanted to use $u$-substitution, you could set $x = \sin u$, which transforms the integral into $\int_0^{\pi/2} du$, giving the same result.

Step-by-Step Solution

1. Identify the antiderivative The integrand $\frac{1}{\sqrt{1-x^2}}$ is the derivative of $\sin^{-1} x$ (also written as $\arcsin x$). So the indefinite integral is:

2. Apply the limits of integration Using the Fundamental Theorem of Calculus:

3. Evaluate the inverse sine values $\sin^{-1}(1) = \frac{\pi}{2}$ (since $\sin(\frac{\pi}{2}) = 1$) $\sin^{-1}(0) = 0$ (since $\sin(0) = 0$) …

Method: Definite integral giving an inverse sine

Recognise $\int\dfrac{dx}{\sqrt{1-x^{2}}}=\sin^{-1}x$, then evaluate across the limits.

Steps

Step 1: Match the standard form. $\dfrac{1}{\sqrt{1-x^{2}}}$ integrates to $\sin^{-1}x$ (not a logarithm).

Step 2: Bracket it.

$\int_{a}^{b}\frac{dx}{\sqrt{1-x^{2}}}=\big[\sin^{-1}x\big]_{a}^{b}.$ …

Common Mistakes

Mistake 1: Confusing $\dfrac{1}{\sqrt{1-x^{2}}}$ with $\dfrac{1}{1+x^{2}}$ (arctan).

Why it's wrong: the radical-in-denominator form is arcsine; the $1+x^{2}$ form is arctangent. Correct approach: $\sqrt{1-x^{2}}\Rightarrow\sin^{-1}x$.

Mistake 2: Using $\sin^{-1}1=\pi$ or $90$ (degrees).

Why it's wrong: $\sin^{-1}1=\dfrac{\pi}{2}$ in radians. Correct approach: work in radians for calculus. …