Question

integrate s4^s ds

Answer 1

\[\large \int\limits_{}^{}s*4^{s} ds\]

Answer 2

i think it is integration by parts u =s, dv= 4^s ds, du = ds, but what is v

Answer 3

4^s / ln(4)

\[\int\limits_{}^{}a^{x} = \frac{a^{x}}{\ln(a)}\]

Answer 4

yes you're right to use integration by parts

Answer 5

final answer is 4^s/ln4(s-1/ln4)+c

Answer 6

thanks, im doing web assign right now, and i keep getting stuck

Answer 7

hmm i get something a little different:

\[\large \frac{s*4^{s}}{\ln(4)} -\frac{4^{s}}{(\ln(4))^{2}}\]

Answer 8

that is the same, mine is just simplified, i took out the s/ln4

Answer 9

how do you draw your equations so nice?

Answer 10

how to integrate e^(−θ) cos 4θ dθ

Answer 11

oh i see, i didn't read your answer right

use the equation button, frac{}{} allows you to write nice fractions

Answer 12

to write exponents....x^{ }

Answer 13

\[e^{-\theta} cos4\theta d\theta\]

Answer 14

integration by parts again...here you will have to do it twice

u = e^-x , dv = cos 4x
du = -e^-x, v = 1/4 sin4x

Answer 15

The 2nd time should look like this:

u = e^-x , dv = sin 4x
du = -e^-x, v = -1/4 cos 4x

Answer 16

resulting in:
\[\int\limits_{}^{}e^{-x}\cos(4x) = \frac{1}{4}e^{-x}\sin(4x)-\frac{1}{16}e^{-x}\cos(4x)-\frac{1}{16}\int\limits_{}^{}e^{-x}\cos(4x)\]

Answer 17

then notice the integrals are exactly the same, so think combining like terms
add 1/16 integral to other side
then all you have to do is divide by a constant

Answer 18

do you follow?

Answer 19

? maybe, but the answer cannot contain an integral

Answer 20

thats right, so imagine you treat the integrals like variables and combine like terms
move it over to other side where the original integral is

Answer 21

awesome, thanks!

Answer 22

your welcome

here is a great resource for checking your work
http://www.wolframalpha.com/input/?i=integrate+e%5E-x+*+cos%284x%29+dx

Answer 23

I appreciate your help, I hope you have a wonderful day