Q632: $\int_{0}^{\pi/2}\frac{\sin~x-\cos~x}{1+\sin~x~\cos~x}dx$ ...

Class CBSE Class 12 Mathematics Definite Integrals Q #632

KNOWLEDGE BASED

APPLY

1 Marks 2024 AISSCE(Board Exam) MCQ SINGLE

$\int_{0}^{\pi/2}\frac{\sin~x-\cos~x}{1+\sin~x~\cos~x}dx$ is equal to:

(A) $\pi$

(B) Zero (0)

(D) $\frac{\pi^{2}}{4}$

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Correct Answer: B

AI Tutor Explanation

Step-by-Step Solution

Let $I = \int_{0}^{\pi/2}\frac{\sin~x-\cos~x}{1+\sin~x~\cos~x}dx$ ----(1)

Using the property $\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx$, we have: $I = \int_{0}^{\pi/2}\frac{\sin(\frac{\pi}{2}-x)-\cos(\frac{\pi}{2}-x)}{1+\sin(\frac{\pi}{2}-x)~\cos(\frac{\pi}{2}-x)}dx$ $I = \int_{0}^{\pi/2}\frac{\cos~x-\sin~x}{1+\cos~x~\sin~x}dx$ ----(2)

Adding equations (1) and (2): $2I = \int_{0}^{\pi/2}\frac{\sin~x-\cos~x}{1+\sin~x~\cos~x}dx + \int_{0}^{\pi/2}\frac{\cos~x-\sin~x}{1+\cos~x~\sin~x}dx$ $2I = \int_{0}^{\pi/2}\frac{\sin~x-\cos~x + \cos~x - \sin~x}{1+\sin~x~\cos~x}dx$ $2I = \int_{0}^{\pi/2}\frac{0}{1+\sin~x~\cos~x}dx$ $2I = 0$ $I = 0$

Correct Answer: 0

AI Suggestion: Option B

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Pedagogical Audit

Bloom's Analysis: This is an APPLY question because the student needs to apply the properties of definite integrals to solve the problem. Specifically, they need to use the property that $\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx$ and then manipulate the resulting integral to find the solution.

Knowledge Dimension: PROCEDURAL

Justification: The question requires the student to follow a specific procedure involving the application of a definite integral property and algebraic manipulation to arrive at the solution.

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

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