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Evaluate the following integrals. Show your work.

Calculus1
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1/cos^2(x) is sec^2(x) and the integral of that is tan(x)
\[\int\limits_{0}^{2}x \sqrt{9-2x^2}dx\]
\[\frac{ d }{ dx }\int\limits_{x^2}^{sinx}\sqrt{1+t^2}dt\]

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Other answers:

no, just substitute, then everything falls out
use calculator so easy
I need to do it without @best_mathematician
ok here...
1 Attachment
integral and derivative cancel each other out
now u know wht to do right @Jgeurts
yes thanks BM
@modphysnoob show me please
\[\frac{ d }{ dx }\int\limits_{x^2}^{sinx}\sqrt{1+t^2}dt\] 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
f(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)
now the bottom part f(x^2)2x=sqrt[1+x^2]2x
sqrt[1+ sin^2(x)]* cos(x)-sqrt[1+x^2]2x
question?
Im formulating my question, haha
ok
so the d/dx means derive the interval?
yes, since we are taking a function integrate it and derive , it undoes the integration
haha great! i get it, thank you, that one was super hard!
I can show you an easy example if you like?
@modphysnoob sure that would be great, im learning for my final :)
let's to a problem like yours but with easier integrand |dw:1367723123193:dw|
so we would derive sinx to cosx and x2 to 2x
then plug in for t and subtract them from each other right?
so cosx-2x?
so 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)
which is cos(x) sin(x)
i see
let's do botttom part d/dx( F (x^2) = 2x f( x^2) 2x x^2= 2x^3
so cos(x) sin(x)-2x^3
this 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
got it?
Yes thats great! thank you so much! @modphysnoob

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