Question

Hello, Can anyone help me with my Calculus question?

Answer 1

shoot

Answer 2

ok :)

Answer 3

\[f ' (x) = 4\sin2x-\sec ^{2}x\]

Answer 4

Find f (x) when f '(x) is given

Answer 5

I ask you shoot and then call other for help you hahahaha...

Answer 6

just take anti derivative or int of that stuff, done

Answer 7

lolz

Answer 8

is it not that?

Answer 9

its integral

Answer 10

yep. the same meaning,

Answer 11

oh

Answer 12

so if you dont mind can you just explain it to me step by step bcoz what ever my proffessor taught I didnt get that

Answer 13

yeah :(

Answer 14

explain or give you the stuff how to do it?

Answer 15

I m actually stuck with Sin and sec thing

Answer 16

now, take int the function, \[\int\4sin 2x - sec^2 x\]

Answer 17

separate into 2 parts

Answer 18

so 4 sin 2x will be 4cos2x?

Answer 19

nope. it's 4 sin 2x not = 4 cos 2x

Answer 20

do you know how to take int of those stuff?

Answer 21

\[\int\4sin2x dx\]= \[4\int\2sinxcosx dx\]

Answer 22

=\[8\int\sinxcosx dx\]

Answer 23

let u = sinx ---> du = cosxdx
substitute into the int. you have \[8\int\udu\]

Answer 24

= ? I don't know, you do next

Answer 25

and then add with the second part which is \[\int\sec^2 xdx\] it has formula to do.
Me done here. ok?

Answer 26

hey, where are you? tell me whether you understand or not???