Question

This integral though

Answer 1

\[\int\limits_{0}^{\infty} \frac{ e ^{-x \sqrt{3}} }{ x }(1-sinx)(1+2sinx-\cos2x)dx\]

Answer 2

\[\int\limits_{0}^{\infty}\frac{ e ^{-x \sqrt{3}} }{ x }(1-sinx)*2(sinx+\frac{ 1-\cos2x }{ 2 })dx\]

Answer 3

\[2\int\limits_{0}^{\infty}\frac{ e ^{-x \sqrt{3}} }{ x }(1-sinx)(sinx+\sin^2x)dx\]

\[2\int\limits_{0}^{\infty}\frac{ e ^{-x \sqrt{3}} }{ x }(1-sinx)sinx(1+sinx)dx\]

Answer 4

\[2\int\limits_{0}^{\infty}\frac{ e ^{-x \sqrt{3}} }{ x }sinx(1-\sin^2x)dx\]

Answer 5

\[2\int\limits_{0}^{\infty}\frac{ e ^{-x \sqrt{3}} }{ x }sinxcos^2xdx\]

\[\int\limits_{0}^{\infty}\frac{ e ^{-x \sqrt{3}} }{ x }(1-sinx)(2sinxcosx)cosxdx\]

Answer 6

\[\int\limits_{0}^{\infty}\frac{ e ^{-x \sqrt{3}} }{ x }1/2(\sin(2x+x)+\sin(2x-x))dx\]

Make sure you jump in, if you see a mistake.

Answer 7

\[1/2\int\limits_{0}^{\infty}\frac{ e ^{-x \sqrt{3}} }{ x }(\sin3x+sinx)dx\]

\[1/2[\int\limits_{0}^{\infty}\frac{ e ^{-x \sqrt{3}} }{ x }\sin3xdx+\int\limits_{0}^{\infty}\frac{ e ^{-x \sqrt{3}} }{ x }sinxdx]\]

Answer 8

Now what?

Answer 9

Would double integration method work

Answer 10

my guess is parts.

Answer 11

all i can think of at this moment is putting a t in place of sqrt(3) and calling the value of the integral I(t)... then differentiation under the integral would eliminate the weird x because d/dt (e^(-xt)) = -x e^(-xt). the e^(ax) sin bx would be a more well known integral to evaluate, then integrate back for I(t)...

there may be a cleaner way to do this though, or maybe this doesn't even check out at the very end?

Answer 12

If you solve for \[
\int e^{-x\sqrt 2}x^{-1}\sin(ax)\;dx
\]then you solve both integrals.

Answer 13

Halfway down page 3, "damped sine integral". http://www.math.uconn.edu/~kconrad/blurbs/analysis/diffunderint.pdf

Answer 14

When doing parts, we generally let \(dv = e^{ax}\)

Answer 15

I mean \(dv=e^{ax}dx\) and \(u=\) the rest

Answer 16

\[\int\limits_{0}^{\infty}\int\limits_{0}^{\infty}e ^{-x \sqrt{3}}sinx*e ^{-xy}dydx\]

Answer 17

Nvm that's tedious, I think Leibiniz integral rule would be the easiest.

Answer 18

Final answer for those of you who are still wondering: pi/6 :)

Answer 19

that looks dirty

Answer 20

I got pi/4 honestly. Are you sure?

Answer 21

I think you're right Kai

Answer 22

What method did you use?

Answer 23

http://openstudy.com/study#/updates/532fd3b5e4b0593075ea3823

Answer 24

\[\int\limits_{0}^{\infty}e^{-tx}\frac{\sin(\lambda x)}{x} dx=\frac{\pi}{2}-\tan^{-1}(\frac{t}{\lambda})\] I just used that link above and followed the reasoning with an extra little parameter lambda.

Answer 25

Ah, I see, well done :P

Answer 26

pee poo hav any more @iambatman ?