Solve ∫dx/√(1-x²) — inverse trigonometric integral

easy CBSE JEE-MAIN • NCERT Class 12 • 2 min read
Tags Integrals

Question

Evaluate ∫dx1−x2\displaystyle\int \frac{dx}{\sqrt{1 - x^2}}.

(NCERT Class 12, Chapter 7 — Integrals)

Solution — Step by Step

This is a standard integral that directly gives an inverse trigonometric function. We need to recall:

∫dx1−x2=sin⁡−1(x)+C\int \frac{dx}{\sqrt{1 - x^2}} = \sin^{-1}(x) + C

This is valid for ∣x∣<1|x| < 1 (the domain where 1−x2\sqrt{1 - x^2} is real and positive).

Let’s confirm: ddx[sin⁡−1(x)]=11−x2\frac{d}{dx}[\sin^{-1}(x)] = \frac{1}{\sqrt{1 - x^2}}

Since the derivative of sin⁡−1(x)\sin^{-1}(x) gives us exactly the integrand, we’ve confirmed:

∫dx1−x2=sin⁡−1(x)+C\boxed{\int \frac{dx}{\sqrt{1 - x^2}} = \sin^{-1}(x) + C}

Since sin⁡−1(x)+cos⁡−1(x)=π2\sin^{-1}(x) + \cos^{-1}(x) = \frac{\pi}{2}, we can also write:

∫dx1−x2=−cos⁡−1(x)+C′\int \frac{dx}{\sqrt{1 - x^2}} = -\cos^{-1}(x) + C'

where C′=C+π/2C' = C + \pi/2. Both answers are correct — they differ by a constant.

Why This Works

The connection between sin⁡−1(x)\sin^{-1}(x) and this integral comes from the substitution x=sin⁡θx = \sin\theta. Then dx=cos⁡θ dθdx = \cos\theta \, d\theta and 1−x2=cos⁡θ\sqrt{1 - x^2} = \cos\theta:

∫cos⁡θ dθcos⁡θ=∫dθ=θ+C=sin⁡−1(x)+C\int \frac{\cos\theta \, d\theta}{\cos\theta} = \int d\theta = \theta + C = \sin^{-1}(x) + C

This substitution is the origin of the formula. Once you’ve derived it once, you can use it as a standard result forever.

Alternative Method — Generalised form

For ∫dxa2−x2\displaystyle\int \frac{dx}{\sqrt{a^2 - x^2}}, substitute x=asin⁡θx = a\sin\theta:

∫dxa2−x2=sin⁡−1(xa)+C\int \frac{dx}{\sqrt{a^2 - x^2}} = \sin^{-1}\left(\frac{x}{a}\right) + C

Our problem is the special case where a=1a = 1.

For CBSE and JEE, memorise these three standard integrals cold:

  • ∫dx1−x2=sin⁡−1(x)+C\int \frac{dx}{\sqrt{1-x^2}} = \sin^{-1}(x) + C
  • ∫dx1+x2=tan⁡−1(x)+C\int \frac{dx}{1+x^2} = \tan^{-1}(x) + C
  • ∫dxxx2−1=sec⁡−1(∣x∣)+C\int \frac{dx}{x\sqrt{x^2-1}} = \sec^{-1}(|x|) + C

These appear directly or as building blocks in almost every integration problem involving inverse trig functions.

Common Mistake

Students sometimes write ∫dx1−x2=cos⁡−1(x)+C\int \frac{dx}{\sqrt{1-x^2}} = \cos^{-1}(x) + C without the negative sign. The derivative of cos⁡−1(x)\cos^{-1}(x) is −11−x2\frac{-1}{\sqrt{1-x^2}} (note the minus sign). So the correct alternative form is −cos⁡−1(x)+C-\cos^{-1}(x) + C, not +cos⁡−1(x)+C+\cos^{-1}(x) + C. Using sin⁡−1(x)\sin^{-1}(x) avoids this sign confusion entirely.

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