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Answer: 1

\(1+\sin2x=\sin^2x+\cos^2x+2\sin x\cos x=(\sin x+\cos x)^2\). On \(\left[0,\dfrac{\pi}{4}\right]\), \(\sin x+\cos x\gt0\), so \(\sqrt{1+\sin2x}=\sin x+\cos x\).

\(I=\displaystyle\int_0^{\pi/4}(\sin x+\cos x)dx=\Big[-\cos x+\sin x\Big]_0^{\pi/4}=\left(-\dfrac{1}{\sqrt2}+\dfrac1{\sqrt2}\right)-(-1+0)=1\).

Doubtly AI · 29 Oct 2026 AI-generated · not yet verified Report
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