Question

Find the antiderivative of sin(2t+5) and e^5x+2x-3

Answer 1

\[\Large \int\limits \sin(2t+5)\;dt\]Mmmmm you can do a u-sub if you want, but guessing is easier imo.
What's the anti-derivative of sine?

Answer 2

-cos

Answer 3

\[\Large \int\limits\limits \sin(2t+5)\;dt\quad\approx\quad -\cos(2t+5)+C\]Ok so there is our guess.
If we take the derivative, do we end up with sin(2t+5)?
How much are we off by?

Answer 4

Where you at sammmm? :P
Don't like this method?

Answer 5

sorry lol

Answer 6

you end up with -1/2cos(2t+5)

Answer 7

Yes you do :3

Answer 8

why doesn't anything change in the parentheses?

Answer 9

using substitution you get
u = 2t + 5
du = 2dt which means dt = du/2
hence your equation becomes
\[\int\limits_{}^{}\frac{ \sin(u)}{ 2 }du\]
now integrating with respect to u gives
\[\frac{ -\cos(u) }{ 2 }\]
and finally you get \[\frac{ -\cos(2t+5) }{ 2 }\]

Answer 10

Why doesn't the inner function change?
Hmm taking the derivative should give you an indication as to why D:
Nothing happens to the inner function when we take a derivative right?

It's the same idea with integration :d

Answer 11

so is -cos(2t+5)/2+C the same as -1/2cos(2t+5)+C

Answer 12

The same? :o no...

It's a different method for solving these types of problems.
I find u-subs to be annoying..
So I use a method of advanced guessing, and then you just adjust your solution at the end.
But do whatever works for you :)

Answer 13

Oh sorry, i didn't see the 2 you wrote :)
yes those are the same!

Answer 14

what if I were to find the derivative instead of the antiderivative?