Determine the value of \(\int_0 ^{\frac{\pi}{2} }(-2cos x)dx\)

Explanation

\(\int_0 ^{\frac{\pi}{2}}(-2cos x)dx = [-2sin x + c]_0 ^{\frac{\pi}{2}}\\
=(-2sin\frac{\pi}{2}+c+2sin0-c)\\
=-2sin90+c+2sin0-c\\
=-2(1)+2(0)\\
=-2\)

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