Class CBSE Class 12 Mathematics Definite Integrals Q #1379
KNOWLEDGE BASED
UNDERSTAND
2 Marks 2025 AISSCE(Board Exam) VSA
Evaluate: $\int_{0}^{\frac{\pi}{4}}\sqrt{1+\sin 2x}dx$.
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Detailed Solution

Step 1: Simplify the integrand

We can rewrite $\sin 2x$ using the identity $\sin^2 x + \cos^2 x = 1$. Thus, we have $\sqrt{1 + \sin 2x} = \sqrt{\sin^2 x + \cos^2 x + 2 \sin x \cos x} = \sqrt{(\sin x + \cos x)^2}$. Since $\sin x + \cos x \geq 0$ for $x \in [0, \frac{\pi}{4}]$, we have $\sqrt{(\sin x + \cos x)^2} = |\sin x + \cos x| = \sin x + \cos x$.

Step 2: Evaluate the integral

Now we can evaluate the integral: $$ \int_{0}^{\frac{\pi}{4}} \sqrt{1 + \sin 2x} \, dx = \int_{0}^{\frac{\pi}{4}} (\sin x + \cos x) \, dx $$ $$ = \left[ -\cos x + \sin x \right]_{0}^{\frac{\pi}{4}} $$ $$ = \left( -\cos \frac{\pi}{4} + \sin \frac{\pi}{4} \right) - \left( -\cos 0 + \sin 0 \right) $$ $$ = \left( -\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} \right) - \left( -1 + 0 \right) $$ $$ = 0 - (-1) = 1 $$

Final Answer: 1

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Pedagogical Audit
Bloom's Analysis: This is an UNDERSTAND question because the student needs to understand trigonometric identities and how to apply them to simplify the integrand before evaluating the definite integral.
Knowledge Dimension: PROCEDURAL
Justification: The question requires the student to apply a specific procedure to solve the integral, including using trigonometric identities and applying the fundamental theorem of calculus.
Syllabus Audit: In the context of CBSE Class 12, this is classified as KNOWLEDGE. The question directly tests the student's knowledge of integration and trigonometric identities, which are core concepts in the syllabus.

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