Class CBSE Class 12 Mathematics Definite Integrals Q #1247
KNOWLEDGE BASED
UNDERSTAND
2 Marks 2024 AISSCE(Board Exam) VSA
Evaluate: $\int_{0}^{\pi/2}sin~2x~cos~3x~dx$
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Detailed Solution

Step 1: Apply Trigonometric Identity

We use the trigonometric identity: $sin~A~cos~B = \frac{1}{2}[sin(A+B) + sin(A-B)]$. Applying this to the integrand, we have: $sin~2x~cos~3x = \frac{1}{2}[sin(2x+3x) + sin(2x-3x)] = \frac{1}{2}[sin~5x + sin(-x)] = \frac{1}{2}[sin~5x - sin~x]$

Step 2: Rewrite the Integral

Now, we rewrite the integral using the identity from Step 1: $\int_{0}^{\pi/2}sin~2x~cos~3x~dx = \int_{0}^{\pi/2} \frac{1}{2}[sin~5x - sin~x]~dx = \frac{1}{2}\int_{0}^{\pi/2} (sin~5x - sin~x)~dx$

Step 3: Integrate

We integrate term by term: $\frac{1}{2}\int_{0}^{\pi/2} (sin~5x - sin~x)~dx = \frac{1}{2} \left[ \int_{0}^{\pi/2} sin~5x~dx - \int_{0}^{\pi/2} sin~x~dx \right]$ $= \frac{1}{2} \left[ -\frac{cos~5x}{5} \Big|_{0}^{\pi/2} + cos~x \Big|_{0}^{\pi/2} \right]$

Step 4: Evaluate the Limits

Now, we evaluate the limits: $= \frac{1}{2} \left[ -\frac{cos(5\pi/2)}{5} + \frac{cos(0)}{5} + cos(\pi/2) - cos(0) \right]$ Since $cos(5\pi/2) = cos(\pi/2) = 0$, $cos(0) = 1$, we have: $= \frac{1}{2} \left[ -\frac{0}{5} + \frac{1}{5} + 0 - 1 \right] = \frac{1}{2} \left[ \frac{1}{5} - 1 \right] = \frac{1}{2} \left[ \frac{1-5}{5} \right] = \frac{1}{2} \left[ \frac{-4}{5} \right] = -\frac{2}{5}$

Final Answer: -2/5

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Pedagogical Audit
Bloom's Analysis: This is an UNDERSTAND question because the student needs to understand trigonometric identities and apply them to solve the integral.
Knowledge Dimension: PROCEDURAL
Justification: The question requires the student to apply a specific procedure (integration) and use trigonometric identities to simplify the integrand before integrating.
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 and trigonometric identities, which are core concepts in the syllabus.

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