Evaluate the definite integral: int_0^{pi /2} | vedclass

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(A) To evaluate $I = \int_0^{\pi /2} \cos^2 x \, dx$,we use the trigonometric identity $\cos^2 x = \frac{1 + \cos(2x)}{2}$.Substituting this into the

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$\frac{\pi}{4}$

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(A) To evaluate $I = \int_0^{\pi /2} \cos^2 x \, dx$,we use the trigonometric identity $\cos^2 x = \frac{1 + \cos(2x)}{2}$.Substituting this into the integral:$I = \int_0^{\pi /2} \frac{1 + \cos(2x)}{2} \, dx$$I = \frac{1}{2} \int_0^{\pi /2} (1 + \cos(2x)) \, dx$$I = \frac{1}{2} \left[ x + \frac{\sin(2x)}{2} \right]_0^{\pi /2}$Evaluating at the limits:$I = \frac{1}{2} \left[ (\frac{\pi}{2} + \frac{\sin(\pi)}{2}) - (0 + \frac{\sin(0)}{2}) \right]$Since $\sin(\pi) = 0$ and $\sin(0) = 0$:$I = \frac{1}{2} [\frac{\pi}{2} + 0 - 0] = \frac{\pi}{4}$.

Evaluate the definite integral: $\int_0^{\pi /2} \cos^2 x \, dx$

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