Question

calculus- Integrals with e in it

Answer 1

I need a few minutes to type the equation

Answer 2

i hate this equation thing lol

Answer 3

\[\int\limits_{}^{}e^{7 \theta} \sin(6 \theta) d \theta ?\]

Answer 4

\[\frac{ 1 }{ 7 } (e^(7\theta)) \sin 6\theta - \frac{ 6 }{ 49 } (e^(7\theta)) \cos 6\theta + \frac{ 36 }{ 56 } (e^(7\theta)) - \cos 6\theta\]

Answer 5

how does that =

Answer 6

those are e raised to 7 theta just so you know

Answer 7

i'm very confused what's going on

Answer 8

can you tell me what you tried to integrate

Answer 9

\[\frac{ 1 }{ 85 } (e^(7\theta)) (7 \sin 6 \theta- 6 \cos 6\theta) + C\]

Answer 10

i was trying to figure out how the first equation i posted = the one i just posted. The math there with the e's have me messed up

Answer 11

\[\int\limits e^7\theta \sin 6\theta (d \theta)\]

Answer 12

that is the original integral question

Answer 13

again that is e raised to 7 theta

Answer 14

so you preformed two rounds of integration
and then that results in an equation you have to solve for the integral you are trying to integrate
so like this:
\[\int\limits_{}^{}e^{ax}\sin(bx) dx=\frac{1}{a}e^{ax}\sin(bx)-\int\limits_{}^{}\frac{1}{a}e^{ax}b \cos(bx) dx \\ = \frac{1}{a}e^{ax}\sin(bx)-[ \frac{1}{a^2}e^{ax} b \cos(bx)- \int\limits_{}^{} \frac{1}{a^2}e^{ax} b^2(-\sin(bx)) dx] \\ =\frac{1}{a}e^{ax}\sin(bx)-\frac{b}{a^2}e^{ax}\cos(bx)-\int\limits_{}^{}\frac{b^2}{a^2}e^{ax} \sin(bx) dx \\ \text{ so adding \to both sides } \int\limits \frac{b^2}{a^2}e^{ax}\sin(bx) dx \\ (1+\frac{b^2}{a^2}) \int\limits e^{ax}\sin(bx) dx=\frac{1}{a}e^{ax}\sin(bx)-\frac{b}{a^2}e^{ax}\cos(bx) \\ \frac{a^2+b^2}{a^2} \int\limits e^{ax} \sin(bx) dx=\frac{1}{a}e^{ax}\sin(bx)-\frac{b}{a^2}e^{ax}\cos(bx) \\ \int\limits e^{ax} \sin(bx) dx= \frac{a^2}{a^2+b^2} (\frac{1}{a}e^{ax}\sin(bx)-\frac{b}{a^2}e^{ax} \cos(bx)) +C \\\]
so you had a=7 and b=6 right...
\[\int\limits_{}^{}e^{7x}\sin(6x) dx=\frac{49}{49+36}(\frac{1}{7}e^{7x}\sin(7x)-\frac{6}{49}e^{7x}\cos(6x)) +C \\ \frac{49}{85}(\frac{1}{7} e^{7x} \sin(6x)-\frac{6}{49} e^{7x} \cos(6x))+C\]

one more step to get that answer:
distribute the 49

Answer 15

\[\frac{1}{85}(\frac{49}{7} e^{7x}\sin(6x)-\frac{49(6)}{49} e^{7x}\cos(6x))+C \\ \frac{1}{85} (7 e^{7x} \sin(6x)-6 e^{7x}\cos(6x))+C\]

Answer 16

I was trying to make sense of what you put but I couldn't ...

Answer 17

hold on... ill show you a pic of the question... its much easier that way. give me one minute

Answer 18

i used integration by parts...

Answer 19

u = sin 6 theta du= 6 cos theta v= e^ 7theta / 7 dv= e^7theta dtheta

Answer 20

that is the problem...what i have of it at least

Answer 21

Ok let me see I'm going to start from this line:
\[\int\limits_{}^{}e^{7x} \sin(6x) dx=\frac{1}{7}e^{7x} \sin(6x) \color{red}- \frac{6}{7}(\frac{1}{7} \cos(6x)e^{7x}+\frac{6}{7} \int\limits e^{7x} \sin(6x) dx)\]
I have a question about this line
why is it - right there
when the sign in front of the integral in the previous step is +

Answer 22

lemme look at it

Answer 23

oh and also for some reason cos changed to sin

Answer 24

but I think it was suppose to be sin from the steps before it

Answer 25

the deriv of cos is -sin....right?

Answer 26

and the deriv of sin is cos?

Answer 27

before the 2nd time integrations by parts started there was a - sign in front of the integral. and inside the integral is a negative6... i figured the minus sign and the negative 6 would make a positive 6

Answer 28

\[I=\int\limits \sin(6x) e^{7x} dx \\ I =\sin(6x)\frac{1}{7} e^{7x}-6 \int\limits_{}^{}\frac{1}{7}e^{7x}\cos(6x) dx \\ I=\frac{1}{7}e^{7x} \sin(6x)-\frac{6}{7} \int\limits e^{7x} \cos(6x) dx \\ I =\frac{1}{7}e^{7x} \sin(6x)-\frac{6}{7}[ \frac{1}{7} e^{7x} \cos(6x)+6\int\limits_{}^{} \frac{1}{7} e^{7x} \sin(6x) dx ] \\ I=\frac{1}{7}e^{7x} \sin(6x)-\frac{6}{49} e^{7x} \cos(6x)-\frac{36}{49} \int\limits e^{7x} \sin(6x) dx\]

Answer 29

\[I=\frac{1}{7}e^{7x} \sin(6x)-\frac{6}{49} e^{7x}\cos(6x)-\frac{36}{49} I\]

Answer 30

solve the equation for I

Answer 31

like do you understand this so far
and is this what I was meant to see?

Answer 32

that is where I got yes

Answer 33

i dont understand how 1/85 occurs

Answer 34

oh okay that's good then

Answer 35

\[\frac{36}{49}I+I=\frac{1}{7}e^{7x} \sin(6x)-\frac{6}{49} e^{7x} \cos(6x)\]

Answer 36

we need to combine fractions
and I think this is the part then you got stuck at

Answer 37

because this is where 1/85 comesinto play I believe

Answer 38

maybe

Answer 39

\[\frac{36}{49}I+\frac{49}{49}I=\frac{1}{7}e^{7x} \sin(6x)-\frac{6}{49} e^{7x} \cos(6x)\]

Answer 40

36+49=85

Answer 41

\[\frac{85}{49} I =\frac{1}{7} e^{7x} \sin(6x)-\frac{6}{49} e^{7x} \cos(6x)\]

Answer 42

so there is your 85

Answer 43

\[\frac{49}{85} \cdot \frac{85}{49} I =\frac{49}{85} (\frac{1}{7} e^{7x} \sin(6x)-\frac{6}{49} e^{7x} \cos(6x))\]

Answer 44

now distribute that 49 on top on the right

Answer 45

\[\frac{ 1 }{ 85 } e(^7\theta)[7\sin(6\theta)- 6\cos(6\theta)] \]

Answer 46

that is the final answer

Answer 47

that is what it told me

Answer 48

we can get that exact answer by distributing the 49 on top
and factoring out the e^(7x)'s

Answer 49

OOOHH i seee

Answer 50

\[\frac{49}{85} \cdot \frac{85}{49} I =\frac{49}{85} (\frac{1}{7} e^{7x} \sin(6x)-\frac{6}{49} e^{7x} \cos(6x)) \\ I=\frac{1}{85}(\frac{49}{7}e^{7x}\sin(6x)-\frac{49(6)}{49} e^{7x} \cos(6x)) \\ I=\frac{1}{85} e^{7x}(7 \sin(6x)-6\cos(6x))\]

Answer 51

it almost looks like in one of your steps you integrate e^(7x )and got 1/8 *e^(7x)

Answer 52

i don't know it gets a little iffy when I start reading your paper towards the end

Answer 53

but anyways the main thing is you get it now

Answer 54

and if you didn't you would tell me right?

Answer 55

I get it BETTER now! it seems like im always fine until the very end

Answer 56

I will review the way you did it again and rewrite me...when i write it out in step by step formation, I sinks in better

Answer 57

mine

Answer 58

you can also look at the thing I did in terms of a and b instead of 6 and 7

Answer 59

like it is a more general example

Answer 60

and it involves less arithmetic

Answer 61

just algebra

Answer 62

thats where i mess up the most... all the arithmetic

Answer 63

the e's confuse me lol

Answer 64

I wish I could say ignore it

Answer 65

me too

Answer 66

either way you were of great help as usual!

Answer 67

ah how sweet