Evaluate the integral \(\int^{\frac{\pi}{4}}_{\frac{\pi}{12}} 2 \cos 2x \mathrm {d} x\)
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Correct Answer: Option C
Explanation:
\(\int_{\frac{\pi}{12}} ^{\frac{\pi}{4}} 2 \cos 2x \mathrm {d} x\)
= \([\frac{2 \sin 2x}{2}]|_{\frac{\pi}{12}} ^{\frac{\pi}{4}}\)
= \(\sin 2x |_{\frac{\pi}{12}} ^{\frac{\pi}{4}}\)
= \(\sin 2(\frac{\pi}{4}) - \sin 2(\frac{\pi}{12})\)
= \(\sin \frac{\pi}{2} - \sin \frac{\pi}{6}\)
= \(1 - \frac{1}{2} = \frac{1}{2}\)
\(\int_{\frac{\pi}{12}} ^{\frac{\pi}{4}} 2 \cos 2x \mathrm {d} x\)
= \([\frac{2 \sin 2x}{2}]|_{\frac{\pi}{12}} ^{\frac{\pi}{4}}\)
= \(\sin 2x |_{\frac{\pi}{12}} ^{\frac{\pi}{4}}\)
= \(\sin 2(\frac{\pi}{4}) - \sin 2(\frac{\pi}{12})\)
= \(\sin \frac{\pi}{2} - \sin \frac{\pi}{6}\)
= \(1 - \frac{1}{2} = \frac{1}{2}\)