Q1292: Evaluate: $\int_{0}^{\frac{\pi^{2}}{4}}\frac{sin\sqrt{x}}{\s...

Class CBSE Class 12 Mathematics Definite Integrals Q #1292

KNOWLEDGE BASED

APPLY

2 Marks 2024 AISSCE(Board Exam) VSA

Evaluate: $\int_{0}^{\frac{\pi^{2}}{4}}\frac{sin\sqrt{x}}{\sqrt{x}}dx$

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AI Tutor Explanation

Detailed Solution

Step 1: Substitution

Let $t = \sqrt{x}$. Then, $t^2 = x$, and differentiating both sides with respect to $x$, we get $2t \, dt = dx$. Also, we need to change the limits of integration. When $x = 0$, $t = \sqrt{0} = 0$. When $x = \frac{\pi^2}{4}$, $t = \sqrt{\frac{\pi^2}{4}} = \frac{\pi}{2}$.

Step 2: Rewrite the integral

Substituting $t = \sqrt{x}$ and $dx = 2t \, dt$, the integral becomes: $$ \int_{0}^{\frac{\pi^{2}}{4}}\frac{\sin\sqrt{x}}{\sqrt{x}}dx = \int_{0}^{\frac{\pi}{2}}\frac{\sin t}{t} (2t \, dt) = 2\int_{0}^{\frac{\pi}{2}} \sin t \, dt $$

Step 3: Evaluate the integral

Now, we evaluate the integral: $$ 2\int_{0}^{\frac{\pi}{2}} \sin t \, dt = 2[-\cos t]_{0}^{\frac{\pi}{2}} = 2\left[-\cos\left(\frac{\pi}{2}\right) - (-\cos(0))\right] = 2[0 - (-1)] = 2(1) = 2 $$

Final Answer: 2

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Pedagogical Audit

Bloom's Analysis: This is an APPLY question because the student needs to apply the technique of substitution to solve the definite integral.

Knowledge Dimension: PROCEDURAL

Justification: The question requires the student to apply a specific procedure (u-substitution) to evaluate the definite integral.

Syllabus Audit: In the context of CBSE Class 12, this is classified as KNOWLEDGE. The question directly tests the student's knowledge of integration techniques, specifically substitution, as covered in the textbook.