Intergal current solution question

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here is the question and solution:

http://i40.tinypic.com/2jd0xnd.gif

its illegal to write an interval t and variable dt

and i can't see how they get their expression.
i got a different result
<br /> dq=\int_{0}^{t}Idt&#039;=It-0<br />

even if i substitute in I the expression that i was given
i still get a different result
??
 
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You assumed I was constant. The exercise, however, gives you an I that depends on t. You need to integrate that function. What result do you get when using that expression?
 
worked thanks :)
 
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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