krypton
y=d/dx integralcosx to 0 t/(1+t)dt
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krypton
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\[d/dx \int\limits_{0}^{cosx} \frac{ t }{ 1+t } dt\]
Outkast3r09
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\[\frac{cosx}{1+cosx}\]
krypton
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how did u do it?
Zekarias
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@Outkast3r09 I think u made a miss take in your first step (While u made the integral)
Outkast3r09
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the derivative of an integral is simply the integral itself with the variables in other words ... integrals and derivatives cancel eachother out
Outkast3r09
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\[\frac{cosx}{1+cosx} \]is your answer
Zekarias
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It should be\[\cos(x)-\ln(1+\cos(x))\]
Outkast3r09
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|dw:1349804491261:dw|
Zekarias
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Now take the derivative
Outkast3r09
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it shouldn't be anything @ zekarias
Outkast3r09
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@zekarias ... you don't need to do anything
Zekarias
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What u get finally?
krypton
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so the answer is just cosx/1+cosx ?
Outkast3r09
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yes
krypton
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thanks,wolfram gives me the same answer,just wanted to make sure how it was done
Outkast3r09
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it's just simply knowing how an integral work... there is a lot of work to do it zekarias way but if you just know that the
the derivative on an integral will bring you back to the f(x) within the integral... that's all you need to know(This works anytime the lower limit is some real number) because if you think about it....
|dw:1349804918350:dw|
if you take the derivative of f(n)... you'll always get zero
krypton
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oh thanks alot sweedy :)