Integration by trigonometric substitution — ∫dx/√(a²-x²) types

Question

Evaluate $\displaystyle\int \frac{dx}{\sqrt{9 - x^2}}$ using trigonometric substitution. Also, state the general formula for integrals of the type $\displaystyle\int \frac{dx}{\sqrt{a^2 - x^2}}$ .

(NCERT Class 12, Integrals)

Solution — Step by Step

The integrand has $\sqrt{a^2 - x^2}$ with $a = 3$ . Whenever we see $\sqrt{a^2 - x^2}$ , we use the substitution:

x = a\sin\theta = 3\sin\theta

This works because $a^2 - x^2 = a^2 - a^2\sin^2\theta = a^2\cos^2\theta$ , which eliminates the square root cleanly.

If $x = 3\sin\theta$ , then $dx = 3\cos\theta \, d\theta$ .

\sqrt{9 - x^2} = \sqrt{9 - 9\sin^2\theta} = \sqrt{9\cos^2\theta} = 3\cos\theta

The integral becomes:

\int \frac{3\cos\theta \, d\theta}{3\cos\theta} = \int d\theta = \theta + C

Since $x = 3\sin\theta$ , we have $\theta = \sin^{-1}\left(\frac{x}{3}\right)$ .

\boxed{\int \frac{dx}{\sqrt{9 - x^2}} = \sin^{-1}\left(\frac{x}{3}\right) + C}

For the general case:

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

This is a standard result that you should memorise. In exams, you can directly use this formula without redoing the substitution.

Why This Works

The substitution $x = a\sin\theta$ converts the algebraic expression $\sqrt{a^2 - x^2}$ into a clean trigonometric one: $a\cos\theta$ . This works because of the Pythagorean identity $\sin^2\theta + \cos^2\theta = 1$ .

The three standard trig substitutions to remember:

Expression	Substitution	Why
$\sqrt{a^2 - x^2}$	$x = a\sin\theta$	Uses $1 - \sin^2\theta = \cos^2\theta$
$\sqrt{a^2 + x^2}$	$x = a\tan\theta$	Uses $1 + \tan^2\theta = \sec^2\theta$
$\sqrt{x^2 - a^2}$	$x = a\sec\theta$	Uses $\sec^2\theta - 1 = \tan^2\theta$

Alternative Method

If you’ve memorised the standard integral, you can skip the substitution entirely:

Compare $\displaystyle\int \frac{dx}{\sqrt{9 - x^2}}$ with $\displaystyle\int \frac{dx}{\sqrt{a^2 - x^2}}$ .

Here $a^2 = 9$ , so $a = 3$ . Apply the formula directly:

\sin^{-1}\left(\frac{x}{3}\right) + C

For JEE Main, memorise all standard integral results from the NCERT table. Most trig substitution problems in JEE reduce to one of these standard forms after algebraic manipulation (completing the square, factoring). The substitution method is the derivation; the formula is what saves time in the exam.

Common Mistake

The most common error: writing $\displaystyle\int \frac{dx}{\sqrt{a^2 - x^2}} = \cos^{-1}\left(\frac{x}{a}\right) + C$ . While technically valid (since $\sin^{-1}(x/a) + \cos^{-1}(x/a) = \pi/2$ , and the $\pi/2$ gets absorbed into $C$ ), NCERT and most exam answer keys expect $\sin^{-1}(x/a) + C$ as the standard answer. Using $\cos^{-1}$ without justification may lose marks in board exams. Stick to the NCERT convention.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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