Evaluate the definite integral int_{0}^{frac{ | vedclass

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(A) Let $I = \int_{0}^{\frac{\pi}{2}} \cos 2x \, dx$. We know that the antiderivative of $\cos 2x$ is $\frac{\sin 2x}{2}$. By the Second Fundamental T

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(A) Let $I = \int_{0}^{\frac{\pi}{2}} \cos 2x \, dx$. We know that the antiderivative of $\cos 2x$ is $\frac{\sin 2x}{2}$. By the Second Fundamental Theorem of Calculus,we have: $I = \left[ \frac{\sin 2x}{2} \right]_{0}^{\frac{\pi}{2}}$ $I = \frac{1}{2} \left[ \sin(2 \cdot \frac{\pi}{2}) - \sin(2 \cdot 0) \right]$ $I = \frac{1}{2} [\sin \pi - \sin 0]$ Since $\sin \pi = 0$ and $\sin 0 = 0$,we get: $I = \frac{1}{2} [0 - 0] = 0$.

Evaluate the definite integral $\int_{0}^{\frac{\pi}{2}} \cos 2x \, dx$.

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