What is I 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{\pi}{4}+\ln 2
  2. B. \frac{\pi}{4}-\ln 2
  3. C. \frac{\pi}{4}-\frac{\ln 2}{2} ✓
  4. D. \frac{\pi}{4}+\frac{\ln 2}{2}

Correct Answer: C. \frac{\pi}{4}-\frac{\ln 2}{2}

Explanation

Let J = \int_0^{\pi/2} \frac{\cos x}{\sin x + \cos x + 1}\,dx. By the property \int_0^a f(x)\,dx = \int_0^a f(a-x)\,dx, applying x \to \pi/2 - x makes I = J. Adding them: I+J = \int_0^{\pi/2} \frac{\sin x + \cos x}{\sin x + \cos x + 1}\,dx = \int_0^{\pi/2} \left(1 - \frac{1}{\sin x + \cos x + 1}\right)\,dx. Using the previous result, this equals \frac{\pi}{2} - \ln 2. Since I=J, 2I = \frac{\pi}{2} - \ln 2, which gives I = \frac{\pi}{4} - \frac{\ln 2}{2}.

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