Question

how do in go about integrating : 8sin^2(t+pi/12) dt

Answer 1

does the pi/12 actually affect anything since its a constant?

Answer 2

no ... of course not

Answer 3

rewrite this into a lower power thru trig identities is my hunch

Answer 4

isnt 8 a constant

Answer 5

cant i take it out

Answer 6

8 is yes

Answer 7

the derivative that pops out of the sin^2 = 1 so thats just there to confuse you

Answer 8

man, i hate the fact that i cant reply quickly on this site, it tells me to wait

Answer 9

rewrite this as: u = t + C; du = dt

\[\int 8sin^2(u)du\]

Answer 10

that cleans up the clutter; now do you recall that sin^2(u) = \(\cfrac{1-cos(2u)}{2}\)??

Answer 11

yes i know that half angle formula

Answer 12

use it in place and itll make life easier then

Answer 13

why is pi/12 a constatn

Answer 14

becasue 3.14/12 is just another number that dissapears in the derivative so its pointless to write out while taking the derivetive :)

Answer 15

so you replaced it with C in the u sub

Answer 16

yes, just to keep from having to type it all in there and waste my fingers ....

Answer 17

its still there in the end; just remember it is all :)

Answer 18

now when i replace it at the end, do i flip those fractions

Answer 19

u = t + 2 ; du = dt

u = t + 13/54 ; du = dt

u = t + pi/12 ; du = dt

u = t + C ; du = dt

Answer 20

oh i got

Answer 21

\[\int 8sin^2(u)du\]
\[8\int sin^2(u)du\]
\[8\int (\frac{1}{2}-\frac{cos(2u)}{2})du\]
\[8(\int \frac{1}{2}du-\int \frac{cos(2u)}{2})du\]
\[8\left(\ \frac{1}{2}\int du-\frac{1}{2}\int cos(2u)\right)du\]

Answer 22

oh man i got lost

Answer 23

lol .... most of that is done in the head

Answer 24

i know but , in the end during computation, i was usppose to get a +9 at the end

Answer 25

\[\int8\frac{1-cos(2u)}{2}du\]
\[\int\frac{8}{2}(1-cos(2u))du\]
\[4\int(1-cos(2u))du\]

Answer 26

a +9?

Answer 27

yeah, i dont know either. this is an initial value problem

Answer 28

where s(0)=8

Answer 29

\[4\int(1-cos(2u))du\]
\[4u -2sin(2u)+C\]
\[4u -2sin(2(t+\frac{pi}{12}))+C\]
\[4u -2sin(2t+\frac{pi}{6})+C\]
oh... initial value then, ok

Answer 30

i forgot a "u" :)

\[4t +\frac{pi}{3} -2sin(2t+\frac{pi}{6})+C\]

Answer 31

how come you didnt replace the u in 4u

Answer 32

now when t = 0 we get:

pi/3 -2 sin(30) + C = 8

Answer 33

i forgot, im old and senile :)

Answer 34

right

Answer 35

but how do we get 9?

Answer 36

pi/3 -2/2 + C = 8

pi/3 + C = 9

C = 9 - pi/3 if i did it right

Answer 37

hmm, let me go over it ans see what i get, thanks for the help

Answer 38

http://www.wolframalpha.com/input/?i=integrate+8sin^2%28t%2Bpi%2F12%29+dt

says i got it good

Answer 39

no, i believe you did it right, i just got to get a hold of it myself :)

Answer 40

good luck :)

Answer 41

hey bro, i got it