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Could someone help me out with calculating this integral:

$\int_{0}^{u} 1/(\sqrt{1-x} \sqrt{x-t})\, dx$

I tried integration by part (maybe not far enough) and it did not seem to bring me to a solution. I also used square completion without success.

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    complete the square in the denominator, then trig substitution.2012-12-14

2 Answers 2

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Completing the square should work.

$\int_{0}^{u} \frac{1}{\sqrt{1-x} \sqrt{x-t}}=\int_{0}^{u} \frac{1}{ \sqrt{-x^2+(t+1)x-1}})\, dx =$ $=\int_{0}^{u} \frac{1}{ \sqrt{-x^2+(t+1)x-\frac{t^2+2t+1}{4}+\frac{t^2+2t-3}{4}}}\, dx $

Now, if $t^2+2t-3 <0$ the function is not well defined. Otherwise, setting $\alpha=\sqrt{\frac{t^2+2t-3}{4}}$ you have

$\int_0^u \frac{1}{\sqrt{\alpha^2-(x-\frac{t+1}{2})^2}}$

which can be calculated with the substitution $x-\frac{t+1}{2}=\alpha \sin (v)$. Of course you need to put some extra restrictions on $t$, to make sure that your function is defined on $[0,u]$.

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    Got it . Thanks.2012-12-14
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Try some changes of variable: first $x = 1 - u^2$, then $u = \sqrt{1-t}\; v$...