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$\displaystyle\int_0^{\pi/2} \dfrac{\sin x}{\sin x + \cos x}\, dx = ?$
A$0$
B$\pi/4$
C$\pi/2$
D$1$
Answer & Solution
Correct answer: B. $\pi/4$
King's rule (substitute $x \to \pi/2 - x$): $I = \int_0^{\pi/2} \cos x/(\cos x + \sin x)\, dx$. Add to original: $2I = \int_0^{\pi/2} (\sin x + \cos x)/(\sin x + \cos x)\, dx = \pi/2$. So $I = \pi/4$.
Related questions
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