Possible to take antiderivative of (sin(x)/e^x)dx?

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Homework Statement



There is an integral, according to an online integral calculator, but I cannot obtain it by integration by parts. Is there a way to obtain it without it being given to you?

The derivative in question is [sin(x)/e^x]dx

Homework Equations





The Attempt at a Solution

 
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Apply integration by parts until you arrive at an integral that looks like the one you started with.

Add the integral to both sides to "cancel" it on one side. Divide by 2 (or whatever other number is now multiplying your integral on the left side of your equation).

Make sure to carefully pick your u and dv. If you pick the right dv, it'll make the problem a lot easier to do. (time wise, I don't think the algebra changes much)
 
Or you could rewrite sinx=\frac{e^{ix}-e^{-ix}}{2i}. Then you would just have to integrate exponentials...
 
I get \frac{cosx+sinx}{-2e^x} for zero constant of integration.
 
You could also write sin(x) as Im{e^(ix)}.

EDIT: beaten by chaoseverlasting
 
In other words, you want
\int e^{-x}sin(x)dx

I see nothing difficult about that. You can do it by parts either by letting u= e-x and dv= sin(x)dx or by letting u= sin(x) and dv= e-xdx.

You will need to integrate twice and after the second integration by parts you will get back to
\int e^{-x}sin(x)dx[/itex]<br /> again so you will have an equation to solve for that integral.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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