Question

hey can s/o help me?

Answer 1

fsm r u gonna help me?

Answer 2

they are not the same because you are finding an "anti derivative" so they can differ by a constant

Answer 3

for example one anti derivative of
\[2x\] is
\[x^2\] but so is
\[x^2+3\]

Answer 4

so whatever two answers you get, they will differ by a contstant

Answer 5

ya but that makes no difference since u just add the +C

Answer 6

but the actual answer will it be diff?

Answer 7

right. so in this case you will get two answers that look different, but they will only differ by a constant

Answer 8

did you do it?

Answer 9

nooo lol. i am actually embarrassed to say that

Answer 10

ok you can just about do it in your head

Answer 11

k gonna do it give me a sec

Answer 12

i got theh same answer for both actually

Answer 13

ok i will be quiet and let you do it yourself. but i will give you a hint. when you do you will get in one case something involving sine squared and in the other something involving cosine squared. then you can figure the constant by recalling that
\[\sin^2(x)+\cos^2(x)=1\]

Answer 14

huh didnt get that LOL

Answer 15

Yeah, pretty much what satellite73 has said.

Answer 16

kind of confused as usual

Answer 17

well you shouldn't
\[\int\sin(x)\cos(x)dx\]
\[u=\cos(x), du = -\sin(x)dx,-du=\sin(x)dx\]
\[-\int udu=-\frac{u^2}{2}=-\frac{\sin(x)}{2}\]

Answer 18

oh i see what i did wrong

Answer 19

repeat the process with
\[u=\sin(x),du=\cos(x)dx\] and you will get
\[\frac{\cos^2(x)}{2}\]

Answer 20

then note that your answers differ by a constant.

Answer 21

ohh okk