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Area under the curve $y=\sin 2 x+\cos 2 x$ between $x=0$ and $x=\frac{\pi}{4}$, is
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1 sq. unit
$\begin{aligned}& \text { Required area }=\int_0^{\pi / 4}(\sin 2 x+\cos 2 x) d x \\& =\left[-\frac{\cos 2 x}{2}+\frac{\sin 2 x}{2}\right]_0^{\pi / 4} \\& =\frac{1}{2}\left[-\cos \frac{\pi}{2}+\sin \frac{\pi}{2}+\cos 0-\sin 0\right]=1 s q unit.\end{aligned}$
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