Discussion
The value of \(\int_{0}^{\pi} \frac{\sin x}{1 + \sin x}dx\) is equal to
Answer: D
The value is \(\pi - 2\).
Write \(\frac{\sin x}{1 + \sin x} = 1 - \frac{1}{1 + \sin x}\).
\[ \begin{aligned} \int_{0}^{\pi}\frac{dx}{1 + \sin x} &= \int_{0}^{\pi}\frac{1 - \sin x}{\cos^{2}x}dx \\ &= \left[\tan x - \sec x\right]_{0}^{\pi} = 2 \\ \text{value} &= \pi - 2 \end{aligned} \]
The antiderivative \(\tan x - \sec x\) equals 1 at \(\pi\) and -1 at 0 (it is continuous across \(\frac{\pi}{2}\), with limit 0 there). \(\pi + 2\) adds instead of subtracting the second part.
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