Question

How is it 1/8.... and not 1/4....?
I did this question multiple time but I keep getting the answer starting with 1/4
http://s1220.photobucket.com/albums/dd441/smileesue/?action=view&current=math.jpg

Answer 1

sorry, it's this link:
http://i1220.photobucket.com/albums/dd441/smileesue/math.jpg

Answer 2

I used Integration by Parts btw

Answer 3

is this in the integration by parts section of your book?
cuz I think I see a different way

Answer 4

\[ \sin x \cos x = \frac 12 \sin 2x \] Now we can can use Integration by parts.

Answer 5

@TuringTest: Yes, it's the integration by parts

Answer 6

@TuringTest: What way is that? Please enlighten us :)

Answer 7

that's actually what I meant, but I didn't realize that leads to integration by parts as well :P

Answer 8

...at least I didn't think it through

Answer 9

Okay fair enough :)

Answer 10

\[(1/2)\int\limits_{}^{}e ^{2x}\sin2xdx\]

Answer 11

That's where I started from and worked my integration of parts from there.

Answer 12

and what did you uses for u and dv ?

Answer 13

(not that is matters here)

Answer 14

lol, why am I typing like I'm drunk ?

Answer 15

u =e^(2x) and du = \[2e ^{2x}dx\]

Answer 16

v = -(cos2x)/2 and dv = sin2xdx

Answer 17

so far so good...
but what's the expression you get?

Answer 18

\[(1/2)\left[ e ^{2x}( -\cos2x/2) + \int\limits_{}^{}(\cos2x/2)(2e ^{2x}dx) \right]\]

Answer 19

so then the 2 gets cancelled out in the second term inside the squared brackets, so I end up with \[\int\limits_{}^{}\cos2x(e ^{2x})dx\] as my second term

Answer 20

yes, with the 1/2 still outside the parentheses

Answer 21

Yes, with the 1/2 outside the squared brackets

Answer 22

Let's call our original integral \(I\)
so the you get\[\frac12\int e^{2x}\sin(2x)=I=-\frac14e^{2x}\cos(2x)+\frac12\int e^{2x}\cos(2x)dx\]now we continue with integration by parts....

Answer 23

\[ \int e ^{2x}\sin2x \; dx = - \frac 12 e^{2x} \cos 2x + \int e^{2x} \cos 2x \; dx \]

\[ = - \frac 12 e^{2x} \cos 2x + \frac 12{e^2x} \sin 2x - \int e ^{2x}\sin2x \; dx \]

Answer 24

I'll just let FFM demonstrate

Answer 25

Which gives \[ 2 \int e ^{2x}\sin2x \; dx = \frac 12 e^{2x} (\sin 2x -\cos 2x ) \] \[\implies \int e ^{2x}\sin2x \; dx = \frac 14 e^{2x} (\sin 2x -\cos 2x ) \]

Answer 26

But we need, \[\frac 12 \int e ^{2x}\sin2x \; dx \] so multilply both sides of the above wiht \( \frac 12 \) to get you desired answer.

Answer 27

It's been more than a year I did any serious integration, so @TuringTest please check for any mistake :P

Answer 28

nope you're right on
let me show that I get the same result

Answer 29

It's good. I am guessing h66 forgot the very last step

Answer 30

Oh well I did some yesterday and today though :P all credits goes to OS

Answer 31

because that is what I did.

Answer 32

I realized if you take \( e^{2x} \) as the second function it would not be any less tedious at-least in this case. Note: if it was \( e^x \) taking it as a second function might help.

Answer 33

\[\frac12\int e^{2x}\sin(2x)=I=-\frac14e^{2x}\cos(2x)+\frac12\int e^{2x}\cos(2x)dx\]\[=-\frac14e^{2x}\cos(2x)+\frac14 e^{2x}\sin(2x)dx-\frac12\int e^{2x}\sin(2x)\]which is the same as\[I=-\frac14e^{2x}\cos(2x)+\frac14 e^{2x}\sin(2x)-I\]\[2I=-\frac14e^{2x}\cos(2x)+\frac14 e^{2x}\sin(2x)\]from which the same result follows

Answer 34

but what i don't get is, doesn't \[1/2\int\limits_{}^{}e ^{2x}\sin2xdx + 1/2\int\limits_{}^{}e ^{2x}\sin2xdx = \int\limits_{}^{}e ^{2x}\sin2xdx\] ?

Answer 35

Yes it does.
But you want 1/2 of the integral to match the original

Answer 36

so \[\int\limits_{}^{}e ^{2x}\sin2xdx = -1/4e ^{2x}\cos2x + 1/4e ^{2x}\sin2x\]

Answer 37

Yes, but you want 1/2 of that.

Answer 38

the integral you want is\[I=\frac12\int e^{2x}\sin(2x)dx\]\(not\)\[\int e^{2x}\sin(2x)dx\]

Answer 39

Change it back to sin(x)cos(x) form to see

Answer 40

Oh, I see my mistake now, after I converted it to sinxcosx. Thank you guys so much! :)