Why can't I substitute a Laplace transform into an integration?

hyphagon
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Hi, I was given the attached question and have given my wrong attempt at the answer. I know how to work the answer out (also shown) but I would like to know why my first attempt is wrong.

Many Thanks
 

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First, you calculated u'(t) incorrectly. Second, when you said v=2/s2, you used the definite integral of dv/dt. You need to use the indefinite integral. You can't get rid of the t dependence.
 
Yup, I integrated it by mistake. That now makes it equal zero incidentally.

Awesome, Thanks for for that. Makes sense now.
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

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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