Evaluate \(\int^{\frac{\pi}{2}} _{\frac{-\pi}{2}} cos x dx\)
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Correct Answer: Option C
Explanation:
\(\int^{\frac{\pi}{2}} _{\frac{-\pi}{2}} cos x dx = [sinx]^{\frac{\pi}{2}} _{\frac{-\pi}{2}}\\
=sin\frac{\pi}{2} - sin\frac{-\pi}{2}\)
= sin90 – sin-90
= sin90 – sin270
= 1 – (-1)
= 1+1
= 2
\(\int^{\frac{\pi}{2}} _{\frac{-\pi}{2}} cos x dx = [sinx]^{\frac{\pi}{2}} _{\frac{-\pi}{2}}\\
=sin\frac{\pi}{2} - sin\frac{-\pi}{2}\)
= sin90 – sin-90
= sin90 – sin270
= 1 – (-1)
= 1+1
= 2