Q1343: Evaluate : $\int_{0}^{\pi}\frac{e^{cos~x}}{e^{cos~x}+e^{-cos...

Class CBSE Class 12 Mathematics Definite Integrals Q #1343

KNOWLEDGE BASED

REMEMBER

3 Marks 2024 AISSCE(Board Exam) SA

Evaluate : $\int_{0}^{\pi}\frac{e^{cos~x}}{e^{cos~x}+e^{-cos~x}}d~x$

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AI Tutor Explanation

Detailed Solution

Step 1: Define the integral

Let $I = \int_{0}^{\pi}\frac{e^{\cos x}}{e^{\cos x}+e^{-\cos x}}dx$

Step 2: Use the property of definite integrals

Using the property $\int_{0}^{a}f(x)dx = \int_{0}^{a}f(a-x)dx$, we have $I = \int_{0}^{\pi}\frac{e^{\cos (\pi - x)}}{e^{\cos (\pi - x)}+e^{-\cos (\pi - x)}}dx$

Step 3: Simplify the integral

Since $\cos(\pi - x) = -\cos x$, we get $I = \int_{0}^{\pi}\frac{e^{-\cos x}}{e^{-\cos x}+e^{\cos x}}dx$

Step 4: Add the two expressions for I

Adding the two expressions for $I$, we have $2I = \int_{0}^{\pi}\frac{e^{\cos x}}{e^{\cos x}+e^{-\cos x}}dx + \int_{0}^{\pi}\frac{e^{-\cos x}}{e^{-\cos x}+e^{\cos x}}dx$ $2I = \int_{0}^{\pi}\frac{e^{\cos x}+e^{-\cos x}}{e^{\cos x}+e^{-\cos x}}dx$ $2I = \int_{0}^{\pi}1 dx$

Step 5: Evaluate the integral

$2I = [x]_{0}^{\pi} = \pi - 0 = \pi$

Step 6: Solve for I

$I = \frac{\pi}{2}$

Final Answer: $\frac{\pi}{2}$

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Pedagogical Audit

Bloom's Analysis: This is an REMEMBER question because it requires recalling and applying a standard property of definite integrals and basic trigonometric identities.

Knowledge Dimension: PROCEDURAL

Justification: The question involves applying a specific procedure (using the property of definite integrals) to evaluate the given integral.<\/span>

Syllabus Audit: In the context of CBSE Class 12, this is classified as KNOWLEDGE. It directly tests the student's understanding and application of definite integrals and their properties as covered in the textbook.