Substitution Rule for Integrals: Solving for the Unknown Variable

dilasluis
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Hello! My problem is the following:

Is

\int_a^b f(z) dt = \int_{g(a)}^{g(b)} f(z) \frac{1}{g} dz

?

\frac{dz}{dt} = g

Thank you!
 
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No. Is z a function of t?
 
If z is a an invertible function of t such that dz/dt= g(t), then dz= g(t)dt, dt= (1/g(t))dz, but you cannot have g(t) in the integral with respect to z.
 
z is a function of t, but not explicit, actually

V_z = \frac{dz}{dt}

was the relation from which we took d t = \frac{dz}{V_z}.

V_z = cte
 
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My biggest problem with this question is f(z) in both sides of the equation... and how do I change the integral from left side to the right.
 
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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