Repeating integration by parts

cameuth
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Homework Statement



integrate .5e^(t/50)*sin(t)

Homework Equations


integration by parts
uv-∫vdu

The Attempt at a Solution



I am currently in differential equations and I remember from cal II that I have to keep using the equation above until the integral loops around, then set it equal to something,... but I'm foggy on what it's equal to. Any help would be appreciated. thanks.
 
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Let's call the integral I. When you do integration by parts repeatedly, you eventually end up with

I = stuff + (something) I

At this point, you solve for I.
 
I believe that you should keep integrating by parts until you end up with something that looks like:

∫.5e^(t/50)*sin(t) = SOMETHING - ∫.5e^(t/50)*sin(t)

then you can add ∫.5e^(t/50)*sin(t) to both sides to get:

2*∫.5e^(t/50)*sin(t) = SOMETHING

so that

∫.5e^(t/50)*sin(t) = (1/2)*SOMETHING

EDIT: vela beat me to it :smile:
 
thanks. vela helped, and saladsamurai's elaboration was necessary. You guys are lifesavers.
 
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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