Question

Evaluate the following integrals. Show your work.

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

no, just substitute, then everything falls out

Answer 5

use calculator so easy

Answer 6

I need to do it without @best_mathematician

Answer 7

ok here...

Answer 8

integral and derivative cancel each other out

Answer 9

now u know wht to do right @Jgeurts

Answer 10

yes thanks BM

Answer 11

@modphysnoob show me please

Answer 12

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

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

Answer 13

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)

Answer 14

now the bottom part

f(x^2)2x=sqrt[1+x^2]2x

Answer 15

sqrt[1+ sin^2(x)]* cos(x)-sqrt[1+x^2]2x

Answer 16

question?

Answer 17

Im formulating my question, haha

Answer 18

ok

Answer 19

so the d/dx means derive the interval?

Answer 20

yes, since we are taking a function integrate it and derive , it undoes the integration

Answer 21

haha great! i get it, thank you, that one was super hard!

Answer 22

I can show you an easy example if you like?

Answer 23

@modphysnoob sure that would be great, im learning for my final :)

Answer 24

let's to a problem like yours but with easier integrand