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1/cos^2(x) is sec^2(x) and the integral of that is tan(x)

\[\int\limits_{0}^{2}x \sqrt{9-2x^2}dx\]

\[\frac{ d }{ dx }\int\limits_{x^2}^{sinx}\sqrt{1+t^2}dt\]

no, just substitute, then everything falls out

use calculator so easy

I need to do it without @best_mathematician

ok here...

integral and derivative cancel each other out

yes thanks BM

@modphysnoob show me please

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?

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

got it?

Yes thats great! thank you so much! @modphysnoob