Class CBSE Class 12 Mathematics Integrals Q #1704
KNOWLEDGE BASED
APPLY
1 Marks 2026 AISSCE(Board Exam) MCQ SINGLE
$\int\frac{dx}{\sqrt{25-16x^{2}}}$ is equal to:
(A) $\frac{1}{5}\sin^{-1}4x+C$
(B) $\frac{1}{25}\sin^{-1}16x+C$
(C) $\frac{1}{4}\sin^{-1}\frac{4x}{5}+C$
(D) $\frac{1}{16}\sin^{-1}\frac{4x}{5}+C$
Correct Answer: C

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Detailed Solution

Step 1: Standard Form Identification

The given integral is of the form ∫ dx / √(a² - u²), which evaluates to sin⁻¹(u/a) + C. We need to manipulate the denominator to match this form.

Step 2: Algebraic Manipulation

Factor out 16 from the square root in the denominator: $$ \int \frac{dx}{\sqrt{25 - 16x^2}} = \int \frac{dx}{\sqrt{16(\frac{25}{16} - x^2)}} $$ $$ = \frac{1}{4} \int \frac{dx}{\sqrt{(\frac{5}{4})^2 - x^2}} $$

Step 3: Integration

Using the standard formula ∫ dx / √(a² - x²) = sin⁻¹(x/a) + C, where a = 5/4: $$ = \frac{1}{4} \sin^{-1}\left(\frac{x}{5/4}\right) + C $$ $$ = \frac{1}{4} \sin^{-1}\left(\frac{4x}{5}\right) + C $$

Final Answer: Option (C)

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Pedagogical Audit
Bloom's Analysis: This is an APPLY question because the student must identify the appropriate standard integral formula and perform algebraic manipulation to transform the expression into the required form.
Knowledge Dimension: PROCEDURAL
Justification: The question tests the student's ability to execute a specific mathematical procedure (integration by substitution/manipulation) rather than just recalling a definition.
Syllabus Audit: In the context of CBSE Class 12, this is classified as KNOWLEDGE. It directly assesses the mastery of standard integration formulas found in Chapter 7 (Integrals) of the NCERT textbook.
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