Q19

1 markSection A

Assertion (A) : $\int \frac{1}{\sqrt{9 - x^2}} dx = \sin^{-1}\frac{x}{3} + C$

Reason (R) : $\int \frac{1}{\sqrt{a^2 - x^2}} dx = \sin^{-1}\frac{x}{a} + C$

Q: Assertion (A) : $\int \frac{1}{\sqrt{9 - x^2}} dx = \sin^{-1}\frac{x}{3} + C$ Reason (R) : $\int \frac{1}{\sqrt{a^2 - x^2}} dx = \sin^{-1}\frac{x}{a} + C$

The correct option is **(A)** — both A and R are true and R is the correct explanation of A. **Reason (R):** The standard result is $\int \frac{dx}{\sqrt{a^2-x^2}} = \sin^{-1}(x/a) + C$ — true. **Assertion (A):** Here $9 = 3^2$, so $a = 3$. Applying R directly gives $\int \frac{dx}{\sqrt{9-x^2}} = \sin^{-1}(x/3) + C$ — true, and it follows exactly from R.

Q: How is the formula $\int \frac{dx}{\sqrt{a^2-x^2}}=\sin^{-1}(x/a)+C$ derived?

Substitute $x = a \sin\theta$, so $dx = a \cos\theta \, d\theta$ and $\sqrt{a^2-x^2} = a \cos\theta$. The integral becomes $\int (a \cos\theta)/(a \cos\theta) \, d\theta = \int d\theta = \theta + C = \sin^{-1}(x/a) + C$, since $\theta = \sin^{-1}(x/a)$.

Integration

Integration — Standard Inverse Trigonometric Form

Options

(A)Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).

(B)Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).

(C)Assertion (A) is true, but Reason (R) is false.

(D)Assertion (A) is false, but Reason (R) is true.

Official Answer

The correct option is (A) — both A and R are true and R is the correct explanation of A.

Reason (R): The standard result is $\int \frac{dx}{\sqrt{a^2-x^2}} = \sin^{-1}(x/a) + C$ — true.

Assertion (A): Here $9 = 3^2$ , so $a = 3$ . Applying R directly gives $\int \frac{dx}{\sqrt{9-x^2}} = \sin^{-1}(x/3) + C$ — true, and it follows exactly from R.

standard integralinverse sinesin inverse formulaa squared minus x squaredassertion reasonintegrationantiderivative

Marking Scheme

11 mark: correct option (A) with recognition that $a = 3$ and R is the general form of A.
2No partial marks; the single correct code earns the full mark.

Hint

Match 9 with $a^2$ : here $a = 3$ , then apply the standard $\sin^{-1}(x/a)$ result.

Quick Oral Answer

Both statements are true and the Reason explains the Assertion, because 9 equals 3 squared, so with a equal to 3 the general formula sin inverse x over a becomes sin inverse x over 3.

Analysis & Explanation

This assertion–reason item checks whether a specific integral is a correct instance of a general standard formula.

Reason (R) is true: $\int \frac{dx}{\sqrt{a^2-x^2}} = \sin^{-1}(x/a) + C$ is one of the standard integrals in the syllabus. It can be verified by differentiating: $\frac{d}{dx}[\sin^{-1}(x/a)] = (1/a)\cdot 1/\sqrt{1-x^2/a^2} = 1/\sqrt{a^2-x^2}$ .

Assertion (A) is true: Writing 9 as $3^2$ identifies $a = 3$ . Substituting into R gives exactly $\sin^{-1}(x/3) + C$ .

R correctly explains A: The Assertion is literally the general Reason applied with $a = 3$ , so the general rule is the precise justification for the particular result.

Correct code — (A). Both statements are true and the Reason is the correct explanation.

Exam trap: Do not confuse this with $\int \frac{dx}{a^2+x^2} = (1/a)\tan^{-1}(x/a)$ or $\int \frac{dx}{\sqrt{a^2+x^2}} = \log|x+\sqrt{a^2+x^2}|$ . The minus sign under the root is what forces the $\sin^{-1}$ form.

Common Mistakes

1Confusing the result with $\int \frac{dx}{a^2-x^2}$ (a logarithm) or $\int \frac{dx}{\sqrt{a^2+x^2}}$ (a log form), leading students to mark R as false.
2Recognising both statements as true but wrongly choosing (B), failing to see that A is a direct special case of R.
3Forgetting the constant of integration C, which is part of the correct standard result.

Interesting Facts

The formula $\int \frac{dx}{\sqrt{a^2-x^2}} = \sin^{-1}(x/a)+C$ comes from the trigonometric substitution $x = a\sin\theta$ , which converts the root into a cosine and simplifies the integral completely.

This same integral gives the area related to a semicircle of radius a, connecting inverse trigonometric functions to geometry.

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Frequently Asked Questions

How is the formula $\int \frac{dx}{\sqrt{a^2-x^2}}=\sin^{-1}(x/a)+C$ derived?

Substitute $x = a \sin\theta$ , so $dx = a \cos\theta \, d\theta$ and $\sqrt{a^2-x^2} = a \cos\theta$ . The integral becomes $\int (a \cos\theta)/(a \cos\theta) \, d\theta = \int d\theta = \theta + C = \sin^{-1}(x/a) + C$ , since $\theta = \sin^{-1}(x/a)$ .

How do I decide the value of 'a' in such problems?

Write the constant under the root as a perfect square. Here $9 = 3^2$ , so $a = 3$ . Then the integral fits the pattern $\sqrt{a^2-x^2}$ exactly and the standard $\sin^{-1}(x/a)$ result applies.