Question

Integration by substitution

Answer 1

Question?

Answer 2

\[\int\limits_{?}^{?}\sin(3-t)dt\]

Answer 3

u = 3-t and du = -1

Answer 4

im not sure how to move on from there :/

Answer 5

\[\int\limits_{}^{}\sin(u)du\]

Answer 6

cos(u) + c ?

Answer 7

du = -1 is incorrect.

du = -dt

Try that.

Answer 8

The negative seems to have vanished into the mist.

Answer 9

it is in there, you just can't see it :P

Answer 10

woops, -cos(u) + c

Answer 11

-cos(3-t) + c

Answer 12

hm i am doing something wrong with the negative. the real answer is cos(3-t) + c

Answer 13

oh just kidding haha integral of sin is -cos so it cancels out

Answer 14

got it, thanks !

Answer 15

This is all very confusing. Let's start from the top and get ALL the signs right.

\(\int \sin(3-t)\;dt\)

u = 3-t
du = -dt

\(\int -\sin(u)\;du = \cos(u) + C\)

Finally, \(\cos(3-t) + C\)

Better to get it right on purpose! :-)

Answer 16

Yeah, I forgot about the -dt and put -1 instead, also the integral of sin is -cos instead of cos. Thanks :)