Question:

Evaluate the integral \( \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos x \, dx \)

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For integrals of even functions over symmetric limits, double the integral over half of the interval.
Updated On: Apr 15, 2025
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The Correct Option is A

Solution and Explanation


The integral we need to solve is: \[ \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos x \, dx \] Since \( \cos x \) is an even function (i.e., \( \cos(-x) = \cos(x) \)), the integral of \( \cos x \) over symmetric limits cancels out: \[ \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos x \, dx = 2 \int_0^{\frac{\pi}{2}} \cos x \, dx \] The integral of \( \cos x \) is \( \sin x \), so: \[ 2 [ \sin x ]_0^{\frac{\pi}{2}} = 2 (1 - 0) = 2 \] Thus, the correct answer is 2.
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