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.

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levetogether 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?

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continue thins thread

levetogether 1

· 8y

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 .

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level 2

· 8y

(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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