Evaluating a definite integral when a condition is given

justwild
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Homework Statement


Given that x^{2}f(x)+f(\frac{1}{x})=0, then evaluate \int^{1.5}_{0.6}f(x)dx

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The Attempt at a Solution



tried to replace f(x) using the provided equation...didn't help
 
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Can you elaborate on how you tried replacing f(x)?
 
Have you looked at what happens if you substitute y = 1/x in the integral and the use the equation to substitute for f(1/y)?
(Are you sure you've quoted the bounds correctly? It's not from a lower bound of 0.66666... by any chance?)
 
haruspex said:
Have you looked at what happens if you substitute y = 1/x in the integral and the use the equation to substitute for f(1/y)?

well If I do that I shall be returning to the same problem statement...
 
justwild said:
well If I do that I shall be returning to the same problem statement...

If the lower limit in the integral is 0.6, you cannot answer the question without knowing more about the function f(x). If the lower limit is 0.6666... = 2/3, you can answer the question without knowing more about f(x).

The substitution u = 1/x DOES work if you do it judiciously!
 
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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