Question 191 markSection A

Assertion (A) : ∫19−x2dx=sin⁡−1x3+C\int \frac{1}{\sqrt{9 - x^2}} dx = \sin^{-1}\frac{x}{3} + C

Reason (R) : ∫1a2−x2dx=sin⁡−1xa+C\int \frac{1}{\sqrt{a^2 - x^2}} dx = \sin^{-1}\frac{x}{a} + C

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 ∫dxa2−x2=sin⁡−1(x/a)+C\int \frac{dx}{\sqrt{a^2-x^2}} = \sin^{-1}(x/a) + C — true.


Assertion (A): Here 9=329 = 3^2, so a=3a = 3. Applying R directly gives ∫dx9−x2=sin⁡−1(x/3)+C\int \frac{dx}{\sqrt{9-x^2}} = \sin^{-1}(x/3) + C — true, and it follows exactly from R.

Answer keywords:standard integralinverse sinesin inverse formulaa squared minus x squaredassertion reasonintegrationantiderivative

Marking scheme

  1. 1.1 mark: correct option (A) with recognition that a=3a = 3 and R is the general form of A.
  2. 2.No partial marks; the single correct code earns the full mark.

Hint

Match 9 with a2a^2: here a=3a = 3, then apply the standard sin⁡−1(x/a)\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.

Common mistakes

  1. 1.Confusing the result with ∫dxa2−x2\int \frac{dx}{a^2-x^2} (a logarithm) or ∫dxa2+x2\int \frac{dx}{\sqrt{a^2+x^2}} (a log form), leading students to mark R as false.
  2. 2.Recognising both statements as true but wrongly choosing (B), failing to see that A is a direct special case of R.
  3. 3.Forgetting the constant of integration C, which is part of the correct standard result.

Analysis and explanation

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


Reason (R) is true: ∫dxa2−x2=sin⁡−1(x/a)+C\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: ddx[sin⁡−1(x/a)]=(1/a)⋅1/1−x2/a2=1/a2−x2\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 323^2 identifies a=3a = 3. Substituting into R gives exactly sin⁡−1(x/3)+C\sin^{-1}(x/3) + C.


R correctly explains A: The Assertion is literally the general Reason applied with a=3a = 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 ∫dxa2+x2=(1/a)tan⁡−1(x/a)\int \frac{dx}{a^2+x^2} = (1/a)\tan^{-1}(x/a) or ∫dxa2+x2=log⁡∣x+a2+x2∣\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\sin^{-1} form.

Interesting facts

The formula ∫dxa2−x2=sin⁡−1(x/a)+C\int \frac{dx}{\sqrt{a^2-x^2}} = \sin^{-1}(x/a)+C comes from the trigonometric substitution x=asin⁡θx = 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.

Question details

Q19 of 44 questions

Board
CBSE
Class
Class 12
Subject
Applied Mathematics
Year
2026
Section
Section A
Marks
1
Chapter
Integration
Topic
Integration — Standard Inverse Trigonometric Form
Bloom's level
understand
Syllabus topic
Unit: Calculus — Integration using the standard result for 1a2−x2\frac{1}{\sqrt{a^2-x^2}}
Author
CBSE
Reviewed by
Padhantu Academic Team

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Frequently asked questions

How is the formula ∫dxa2−x2=sin⁡−1(x/a)+C\int \frac{dx}{\sqrt{a^2-x^2}}=\sin^{-1}(x/a)+C derived?

Substitute x=asin⁡θx = a \sin\theta, so dx=acos⁡θ dθdx = a \cos\theta \, d\theta and a2−x2=acos⁡θ\sqrt{a^2-x^2} = a \cos\theta. The integral becomes ∫(acos⁡θ)/(acos⁡θ) dθ=∫dθ=θ+C=sin⁡−1(x/a)+C\int (a \cos\theta)/(a \cos\theta) \, d\theta = \int d\theta = \theta + C = \sin^{-1}(x/a) + C, since θ=sin⁡−1(x/a)\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=329 = 3^2, so a=3a = 3. Then the integral fits the pattern a2−x2\sqrt{a^2-x^2} exactly and the standard sin⁡−1(x/a)\sin^{-1}(x/a) result applies.