Question #8c45d

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Question #8c45d

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Answer 1 · 2016-11-30T18:25:28

$\int_{0}^{1} 2 cos (\frac{π t}{2}) d t = \frac{4}{π}$ $int_0^1 2 cos ((pit)/2) dt=4/pi$

Explanation:

Provide the preparation $\frac{π}{2}$ $pi/2$ inside the integral and $\frac{2}{π}$ $2/pi$ outside the integral.

$\int_{0}^{1} 2 cos (\frac{π t}{2}) d t = (\frac{2}{π}) \cdot 2 \cdot \int_{0}^{1} cos (\frac{π t}{2} \cdot (\frac{π}{2}) d t$ $int_0^1 2 cos ((pit)/2) dt=(2/pi)*2*int_0^1 cos ((pit)/2*(pi/2)dt$

$\int_{0}^{1} 2 cos (\frac{π t}{2}) d t = (\frac{4}{π}) {[sin (\frac{π t}{2})]}_{0}^{1}$ $int_0^1 2 cos ((pit)/2) dt=(4/pi)[sin ((pit)/2)]_0^1$

$\int_{0}^{1} 2 cos (\frac{π t}{2}) d t = \frac{4}{π} [sin ((\frac{π}{2}) (1)) - sin ((\frac{π}{2}) (0))]$ $int_0^1 2 cos ((pit)/2) dt=4/pi[sin ((pi/2)(1))-sin ((pi/2)(0))]$

$\int_{0}^{1} 2 cos (\frac{π t}{2}) d t = \frac{4}{π}$ $int_0^1 2 cos ((pit)/2) dt=4/pi$

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Question #8c45d

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