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NDA 2 2024 Maths, question 97

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.

Q97

What is ∫0π/2dxg(x)\int _{0}^{\pi/2}\frac{dx}{g(x)} equal to?

  1. (a) ln⁡22\frac{\ln 2}{2}
  2. (b) ln⁡24\frac{\ln 2}{4}
  3. (c) ln⁡2\ln 2Correct answer
  4. (d) 2ln⁡22\ln 2

Solution.

Answer: (c) ln⁡2\ln 2

  1. Use t=tan(x/2), with x from 0 to π/2 corresponding to t from 0 to 1.
  2. Then sin x+cos x+1=2(t+1)/(1+t²) and dx=2dt/(1+t²).
  3. The integral becomes ∫₀¹dt/(1+t)=ln2.