How does du replace the circled part of the question in this integral?

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Homework Statement



http://img528.imageshack.us/img528/356/hwhelp.png

Homework Equations



http://img407.imageshack.us/img407/3696/uanddu.png

The Attempt at a Solution


This is the answer I got. I was wondering how du replaces the circled part of the question.

http://img546.imageshack.us/img546/8975/hwanswer.png
 
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I just came up with this information now. Is it right?

http://img851.imageshack.us/img851/3763/uandupart2.png
 
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november1992 said:

Homework Statement



http://img528.imageshack.us/img528/356/hwhelp.png

Homework Equations



http://img407.imageshack.us/img407/3696/uanddu.png

The Attempt at a Solution


This is the answer I got. I was wondering how du replaces the circled part of the question.

http://img546.imageshack.us/img546/8975/hwanswer.png

You have
du = \frac{e^{\sqrt{x}}}{2\sqrt{x}}dx = \frac{1}{2\sqrt{x}e^{-\sqrt{x}}}dx

The part you have circled (and including dx) is 2du.
 
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Thanks, I get it now.
 
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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