Substitute x = a sin θ to evaluate. Result: sin⁻¹(x/a) + C.

Integration Using Trigonometric Substitution — ∫dx/√(a²-x²)

Question

Evaluate:

\int \frac{dx}{\sqrt{a^2 - x^2}}

This is a standard NCERT Class 12 integral that shows up in both board exams and JEE Main. The trick is recognizing the Pythagorean identity hiding inside the square root.

Solution — Step by Step

The expression $a^2 - x^2$ looks like $a^2 - a^2\sin^2\theta = a^2(1 - \sin^2\theta)$ . So we substitute $x = a\sin\theta$ .

This works because we’re replacing a messy square root with a clean trigonometric expression.

Differentiate $x = a\sin\theta$ :

dx = a\cos\theta \, d\theta

\sqrt{a^2 - x^2} = \sqrt{a^2 - a^2\sin^2\theta} = \sqrt{a^2\cos^2\theta} = a\cos\theta

The integral becomes:

\int \frac{a\cos\theta \, d\theta}{a\cos\theta} = \int d\theta = \theta + C

Since $x = a\sin\theta$ , we get $\sin\theta = \frac{x}{a}$ , so $\theta = \sin^{-1}\!\left(\frac{x}{a}\right)$ .

\boxed{\int \frac{dx}{\sqrt{a^2 - x^2}} = \sin^{-1}\!\left(\frac{x}{a}\right) + C}

Why This Works

The substitution $x = a\sin\theta$ is effective because it exploits the Pythagorean identity $\sin^2\theta + \cos^2\theta = 1$ . When we write $a^2 - x^2 = a^2(1 - \sin^2\theta)$ , the square root collapses into $a\cos\theta$ — no square root, no problem.

This is the standard pattern: whenever you see $\sqrt{a^2 - x^2}$ , think $x = a\sin\theta$ . For $\sqrt{a^2 + x^2}$ , use $x = a\tan\theta$ . For $\sqrt{x^2 - a^2}$ , use $x = a\sec\theta$ . Learning these three patterns covers almost every trig substitution question in boards and JEE Main.

For the special case $a = 1$ , the result simplifies to $\sin^{-1}(x) + C$ . This is worth memorising directly since it appears frequently in composite integrals.

Alternative Method — Direct Formula Recognition

Once you’ve derived this result a few times, you recognise the standard form directly. The formula:

\int \frac{dx}{\sqrt{a^2 - x^2}} = \sin^{-1}\!\left(\frac{x}{a}\right) + C

is listed in NCERT’s standard integral table. In board exams, you can apply it directly without re-deriving — just state “using standard result.”

This becomes powerful when handling integrals like $\int \frac{dx}{\sqrt{9 - 4x^2}}$ . Rewrite as $\int \frac{dx}{\sqrt{(3)^2 - (2x)^2}}$ , substitute $u = 2x$ , $du = 2\,dx$ , and apply the formula with $a = 3$ .

Common Mistake

Forgetting to write $\pm a\cos\theta$ , then ignoring the sign.

When we simplify $\sqrt{a^2\cos^2\theta}$ , the result is $|a\cos\theta|$ . For the substitution to work cleanly, we assume $\theta \in [-\pi/2, \pi/2]$ , which means $\cos\theta \geq 0$ , so we can drop the absolute value. Many students skip this assumption entirely and later get confused when the derivation seems “incomplete” in a viva or JEE paper where steps must be justified. Always state the range of $\theta$ when you make the substitution.

Question

Solution — Step by Step

Why This Works

Alternative Method — Direct Formula Recognition

Common Mistake

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