Is this an appropriate assumption

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<br /> \frac{d^2}{dx^2}\,\int_{0}^{x}\Bigg(\int_{1}^{sint}\,\sqrt{1+u^4}\,du\Bigg)\,dt<br />
When solving something like this is it appropriate to look at it (for sake of ease), as just replacing u^4 with \sin{t} then multiplying the original expression by the derivative of \sin{t}?
 
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That upper bound is sint (dont know why it won't show up).
and that's replacing u^4 with sint and the derivative of sint.
 
mateomy said:
<br /> When solving something like this is it appropriate to look at it (for sake of ease), as just replacing u^4 with \sin{t} then multiplying the original expression by the derivative of \sin{t}?
<br /> <br /> What you&#039;re saying sounds vaguely like something you will need to do when you solve the problem correctly. You want to use the fundamental theorem of calculus (twice actually) first. The second time you will also need to use the chain rule, which is what you seem to be trying to say in other words. It&#039;s best to actually use the rules here to take the derivatives rather than try to guess at how things will fit together.
 
Yeah, Fundamental Theorem...exactly what I (wasn't) saying, haha. Thanks. Just clarifying things in my own head; finals in 2 weeks. Thank you for confirming.
 
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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