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On [0, pi/4], the integral of sinxdx=

User RuBiCK
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Answer: The integral of sin(x)dx on the interval [0, pi/4] is:

∫sin(x)dx = -cos(x) + C

where C is the constant of integration.

To evaluate this definite integral on the interval [0, pi/4], we substitute pi/4 for x in the antiderivative and then subtract the value of the antiderivative at x=0:

cos(pi/4) - (-cos(0)) = -(√2/2) - (-1) = 1 - √2/2

Therefore, the value of the integral of sin(x)dx on the interval [0, pi/4] is 1 - √2/2.

User Zhekaus
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