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anonymous
 3 years ago
Evaluate the following integrals. Show your work.
anonymous
 3 years ago
Evaluate the following integrals. Show your work.

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zepp
 3 years ago
Best ResponseYou've already chosen the best response.01/cos^2(x) is sec^2(x) and the integral of that is tan(x)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{0}^{2}x \sqrt{92x^2}dx\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\frac{ d }{ dx }\int\limits_{x^2}^{sinx}\sqrt{1+t^2}dt\]

primeralph
 3 years ago
Best ResponseYou've already chosen the best response.1no, just substitute, then everything falls out

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0use calculator so easy

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I need to do it without @best_mathematician

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0integral and derivative cancel each other out

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0now u know wht to do right @Jgeurts

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@modphysnoob show me please

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\frac{ d }{ dx }\int\limits_{x^2}^{sinx}\sqrt{1+t^2}dt\] after our supposed integration F(Sin(x))  F ( x^2) take dervaitve d/dx of F( Sin(x)) = cos(x) f(sin(x) d/dx of F(x^2) = f (x^2) 2x go back and plug into your funciton

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0f(t)= Sqrt[1+t^2] f(sin(x)= sqrt[1+ sin^2(x)] f(sin(x)*cos(x)= sqrt[1+ sin^2(x)]* cos(x)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0now the bottom part f(x^2)2x=sqrt[1+x^2]2x

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0sqrt[1+ sin^2(x)]* cos(x)sqrt[1+x^2]2x

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Im formulating my question, haha

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so the d/dx means derive the interval?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yes, since we are taking a function integrate it and derive , it undoes the integration

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0haha great! i get it, thank you, that one was super hard!

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I can show you an easy example if you like?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@modphysnoob sure that would be great, im learning for my final :)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0let's to a problem like yours but with easier integrand dw:1367723123193:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so we would derive sinx to cosx and x2 to 2x

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0then plug in for t and subtract them from each other right?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so in this case f(t)= t when we integrate we will get a function which we will call F(t) F(t) is evaluated at F(sin(x))  F( x^2) then taken d/dx of d/dx( F(sin(x))  F( x^2) ) using chain rule d/dx ( F(sin(x)) = cos(x) f( sin(x)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0which is cos(x) sin(x)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0let's do botttom part d/dx( F (x^2) = 2x f( x^2) 2x x^2= 2x^3

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so cos(x) sin(x)2x^3

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0this is easy integrand so we could actually integrate it and then differntiate it to confirm out result. Integrate t t^2/2 plug in limits (sin^2 x)/2  x^4/2 take derivative d/dx cos(x) sin(x) 2x^3

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Yes thats great! thank you so much! @modphysnoob
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