Trigonmetric integrals and substitutions

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Homework Statement


integral of 83/((x^2)(100x^2-121)^(1/2))


Homework Equations


1+(tanx)^2=(secx)^2

(secx)^2-1=(tanx)^2

(x^2-a^2)^(1/2)

The Attempt at a Solution


moved 83 out of the integral tried converting the denominator to sec and then to tan, but not sure if that is right because of the 100.
 
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First substitute 11*u/10=x. Now take the 121 out of the radical and put it with your 83 (after you take the sqrt of course!). The meat of the integral is 1/(u^2*sqrt(u^2-1)). That looks like a sec substitution.
 
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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