girlfriend forobtained a critical part the ns indefinite integral: the consistent the integration. Those indefinite integralns should be
∫ sin 2x dx = −cos 2x + C,
∫ 2 sin x cos x dx = sin2 x + C.
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ns continuous the integrati~ above have the right to it is in any consistent number, Because constants vanish when friend differentiate.
So, what happens when C = 1/2 in the expressitop top −cos 2x + C? you get
−cos 2x + 1/2 = −(cos2 x − sin2 x)/2 + 1/2 = (sin2 x + 1 − cos2 x)/2 = (sin2 x)/2 + (1 − cos2 x)/2 = (sin2 x)/2 + (sin2 x)/2 = sin2 x.
for this reason this 2 different-lookinns expression for ns antiderivative are yes, really the exact same thing—they different only by a constant, 1/2.
Op · 8y
i believed C is an 'arbitrarily constant' not a certain one. Just how should ns understand when to usage a particular constant, ~ above a test for example?
continue thins thread
girlfriend foracquired the continuous in her initially integration. It should reAD (assuming girlfriend combine indigenous 0 to x) : (1-cos(2x))/2.
Which, as can be checked, ins Indeed (sin x)2 .
(assuming you combine native 0 come x)
OP's integrals to be indefinite integrals, no definite integrals. There are no boundns that integration, and also the price should save the constant of integrati~ above ("+ C").
the variable x in this integralns is a fake variable. Also if they were delimited integrals, ins wouldn't do feeling to set ns boundns come it is in 0 to x, Since x is a dummy variable that has no interpretation "top top the outside" of ns integral.
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