Definite Integration
ID: #28 | Difficulty: medium
AI Quick Answer:
The correct option for this defence exam question is (B) which is \pi/4. Detailed proof and explanation are supplied below.
Evaluate the definite integral: I = \int_{0}^{\pi/2} \frac{\sin^{{n}}(x)}{\sin^{{n}}(x) + \cos^{{n}}(x)} dx, where n = 18.
Key Takeaways for NDA & CDS Candidates
- Ensure you recall the standard formulas for Definite Integration during UPSC exam preparation.
- Eliminating obviously incorrect choices based on dimensional analysis or limits saves valuable seconds.
- Expert tip: Written defence mock tests should be practiced weekly with strict negative marking rules.
Correct Answer: (B)
\pi/4
Step-by-Step Explanation
By applying the properties of definite integrals, specifically \int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx, we can show that:
I = \int_{0}^{\pi/2} \frac{\sin^{n}(\pi/2-x)}{\sin^{n}(\pi/2-x) + \cos^{n}(\pi/2-x)} dx = \int_{0}^{\pi/2} \frac{\cos^{n}(x)}{\cos^{n}(x) + \sin^{n}(x)} dx.
Adding the two equations:
2I = \int_{0}^{\pi/2} \frac{\sin^{n}(x) + \cos^{n}(x)}{\sin^{n}(x) + \cos^{n}(x)} dx = \int_{0}^{\pi/2} 1 dx = [x]_{0}^{\pi/2} = \pi/2.
Therefore, I = \pi/4. The value of n (18) does not affect the final answer.
I = \int_{0}^{\pi/2} \frac{\sin^{n}(\pi/2-x)}{\sin^{n}(\pi/2-x) + \cos^{n}(\pi/2-x)} dx = \int_{0}^{\pi/2} \frac{\cos^{n}(x)}{\cos^{n}(x) + \sin^{n}(x)} dx.
Adding the two equations:
2I = \int_{0}^{\pi/2} \frac{\sin^{n}(x) + \cos^{n}(x)}{\sin^{n}(x) + \cos^{n}(x)} dx = \int_{0}^{\pi/2} 1 dx = [x]_{0}^{\pi/2} = \pi/2.
Therefore, I = \pi/4. The value of n (18) does not affect the final answer.
| Question Metadata Summary | |
|---|---|
| Exam Category | UPSC NDA / CDS / AFCAT / Agniveer |
| Subject Unit | NDA Maths Questions |
| Sub-topic / Module | Definite Integration |
| Correct Option | Option (B) |
| Target Year | UPSC Defence Exams 2026 |