What is \int_{0}^{\pi/2}\frac{dx}{g(x)} equal to?

Direction: Consider the following for the two (02) items that follow : Let I=\int_{0}^{\pi/2}\frac{f(x)}{g(x)}\,dx, where f(x)=\sin x and g(x)=\sin x+\cos x+1.

  1. A. \frac{\ln 2}{2}
  2. B. \frac{\ln 2}{4}
  3. C. \ln 2 ✓
  4. D. 2\ln 2

Correct Answer: C. \ln 2

Explanation

Let the integral be K = \int_0^{\pi/2} \frac{1}{\sin x + \cos x + 1}\,dx. Using the half-angle substitution t = \tan(x/2), dx = \frac{2\,dt}{1+t^2}, \sin x = \frac{2t}{1+t^2}, and \cos x = \frac{1-t^2}{1+t^2}. The limits change from 0 \to 1. Thus, K = \int_0^1 \frac{1}{\frac{2t}{1+t^2} + \frac{1-t^2}{1+t^2} + 1} \frac{2}{1+t^2}\,dt = \int_0^1 \frac{2}{2t + 1 - t^2 + 1 + t^2}\,dt = \int_0^1 \frac{2}{2t+2}\,dt = \int_0^1 \frac{1}{t+1}\,dt = [\ln|t+1|]_0^1 = \ln 2.

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