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NDA 2 2024 Maths, question 98

Topic: Integral calculus

Passage · Q97 to Q98

Let I=∫0π/2f(x)g(x)dxI=\int _{0}^{\pi/2}\frac{f(x)}{g(x)} dx, where f(x)=sin⁡xf(x)=\sin x and g(x)=sin⁡x+cos⁡x+1g(x)=\sin x+\cos x+1.

Q98

What is II equal to?

  1. (a) π4+ln⁡2\frac{\pi}{4}+\ln 2
  2. (b) π4−ln⁡2\frac{\pi}{4}-\ln 2
  3. (c) π4−ln⁡22\frac{\pi}{4}-\frac{\ln 2}{2}Correct answer
  4. (d) π4+ln⁡22\frac{\pi}{4}+\frac{\ln 2}{2}

Solution.

Answer: (c) π4−ln⁡22\frac{\pi}{4}-\frac{\ln 2}{2}

  1. The substitution x↦π/2−x shows I also equals the integral of cos x/(sin x+cos x+1).
  2. Add the two forms: 2I=∫₀^(π/2)[1−1/(sin x+cos x+1)]dx=π/2−ln2.
  3. Therefore I=π/4−(ln2)/2.