Integration by parts, help me understand why the integration limits changed.

truman
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Homework Statement


I am doing self-study. I am on problem 5.6 #27 in the Stewart text 3rd E.

I don't understand why the integration limits changed after the given substitution.

The given substitution was:

x=θ^2 dx=2θdθ

Homework Equations


Please see attachment.


The Attempt at a Solution



I understand the substitution, and how theta sq became 2theta d theta. What I don't understand is, why when the x was substituted, the integration limits changed from a square root to no square root.
 

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The original limits of integration are from \theta= \sqrt{\pi/2} to \theta= \sqrt{\pi}. With x= \theta^2 they become x= (\sqrt{\pi/2})^2= \pi/2 and x= (\sqrt{\pi})^2= \pi.
 
Yes, I see. You have to substitute in the variable and then it's squared, removing the square root sign.

Brain fart. Thank you!
 
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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