Solving Incomplete Integral: Easy Steps and Tips | Andishe9

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Hi all,
can anybody complete my answer?(I'm unable to solve the second part of the integral)
problem and answer(incomplete) : www.andishe9.com/integral.bmp[/URL]

thanks and excuse me for english;)
 
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Welcome to PF!

Hamid1 said:
Hi all,
can anybody complete my answer?(I'm unable to solve the second part of the integral)
problem and answer(incomplete) : www.andishe9.com/integral.bmp[/URL]

thanks and excuse me for english;)[/QUOTE]

Hi Hamid1! Welcome to PF! :smile:

[SIZE="1"](have an integral: ∫ :wink:)

your link isn't working …

can you please type out your answer?
 
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Thank you,
I checked the link.It works...
 
ermmm...your very first line is incorrect: your basically claiming that (2x+6)-6=1 which is clearly false.

Try a u-substitution of the form u=x+3 instead.
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

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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