Evaluate the integral of $ cos(2x) $ from $ 0 $ to $ frac{pi}{2} $
Answer 1
Charlotte Davis
To evaluate the integral of $ \cos(2x) $ from $ 0 $ to $ \frac{\pi}{2} $:
$ \int_0^{\frac{\pi}{2}} \cos(2x) \, dx $
Use the substitution $ u = 2x $, then $ du = 2dx $ or $ dx = \frac{1}{2} du $:
$ \int_0^{\frac{\pi}{2}} \cos(2x) \, dx = \frac{1}{2} \int_0^{\pi} \cos(u) \, du $
The integral of $ \cos(u) $ is $ \sin(u) $:
$ \frac{1}{2} \left[ \sin(u) \right]_0^{\pi} $
Evaluate the definite integral:
$ \frac{1}{2} \left( \sin(\pi) – \sin(0) \right) = \frac{1}{2} (0 – 0) = 0 $
Therefore, the final answer is:
$ 0 $
Answer 2
Daniel Carter
To evaluate the integral of $ cos(2x) $ from $ 0 $ to $ frac{pi}{2} $, use the substitution $ u = 2x $:
$ int_0^{frac{pi}{2}} cos(2x) , dx = frac{1}{2} int_0^{pi} cos(u) , du $
The integral of $ cos(u) $ is $ sin(u) $:
$ frac{1}{2} left[ sin(u)
ight]_0^{pi} $
Evaluate the definite integral:
$ frac{1}{2} (0 – 0) = 0 $
Answer 3
Matthew Carter
To evaluate the integral of $ cos(2x) $ from $ 0 $ to $ frac{pi}{2} $, use the substitution $ u = 2x $:
$ frac{1}{2} int_0^{pi} cos(u) , du $
The result is $ 0 $
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