How do I evaluate this integral?

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I don't understand how the book example went from

integral from -1 to 0 of [e^(x).cos(e^x)]dx

to sin(e^x)]evaluate at -1 and 0.

Can someone please explain?
 
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frasifrasi said:
I don't understand how the book example went from

integral from -1 to 0 of [e^(x).cos(e^x)]dx

to sin(e^x)]evaluate at -1 and 0.

Can someone please explain?

Use a substitution of u = e^x.
 
Yes! thank you.
 
frasifrasi said:
Yes! thank you.

no prob.
 
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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