Integral of [1-(e^(t/a))]^2 | LR Circuit Power

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[SOLVED] integral problem

Homework Statement



hi. my problems an integral which i can't solve. the integral is this-

integral of [1-(e^(t/a))]^2 dt

a is a constant. it came when i tried to find the total power through a LR circuit


Homework Equations





The Attempt at a Solution



not sure at all.

[1-(e^(t/a))]^3
----------------
3 [ then? ]
 
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This can be done with substitution:
\int(1-{e}^{t/a})^{2}dt
Let u=t/a and thus du=dt/a, dt={a}{du}.
After expanding the integrand and putting a (the constant) on the outside, it becomes:
{a}\int{e}^{2u}-2e^u+1 du.
Again, if you don't see it already, use substitution on the first part and manually do the other two easy parts.
Let w=2u, dw=2du, \frac{1}{2}dw=du
\frac{1}{2}a\int{e^w}dw=\frac{1}{2}ae^w=\frac{1}{2}ae^{2u}=\frac{1}{2}ae^{2t/a}
The second part is easy:
-a \int{2e^u}du=-2e^u=-2e^{t/a}
The third part is the easiest (I added the constant 'C' here):
a \int du=au+C=a\frac{t}{a}+C=t+C don't forget the substitution!
Summing all of these gives these final answer:
\int(1-{e}^{t/a})^{2}dt=\frac{1}{2}ae^{2t/a}-2e^{t/a}+t+C
 
hey thanks! a lot. how did you get the integral signs? (the big s) and the proper format?
 
hey thanks! a lot. how did you get the integral signs? (the big s) and the proper format?
 
harshasunder said:
hey thanks! a lot. how did you get the integral signs? (the big s) and the proper format?

\LaTeX is what you're looking for go to

https://www.physicsforums.com/showthread.php?t=8997

\int_0^{\infty} e^{x^2}\;dx\rightarrow \int \mid e^{x^2}\mid \lim_{x\rightarrow\infty}=

\sum_{n=-\infty}^\infty \frac{x^n}{n!} = \lim_{n\rightarrow\infty} (1+x/n)^n

You can click on any bit of code and cut and paste it as well. Try these ones.
 
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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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