Discussion
The value of \(I = \int_{0}^{\frac{\pi}{2}} \frac{(sinx+cosx)^{2}}{\sqrt{1+sin2x}}\, dx\) is
Answer: C
The integral equals 2.
Since \(sin2x = 2 sinx cosx\), the denominator is \(\sqrt{sin^{2}x + cos^{2}x + 2sinx cosx} = |sinx + cosx|\), and on \([0, \pi/2]\) both functions are non-negative, so it is simply sinx + cosx.
\[ \begin{aligned} I &= \int_{0}^{\pi/2} (sinx + cosx)\, dx \\ &= \big[-cosx + sinx\big]_{0}^{\pi/2} \\ &= (0 + 1) - (-1 + 0) = 2 \end{aligned} \]
1 is the value of either half of the integral alone, and 0 comes from a sign slip, taking the antiderivative of sinx as +cosx so the two parts cancel.
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