Inverse Tan Function Range: (-pi/2, pi/2)

missmerisha
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Homework Statement



State the range

y=inverse tan( 2x)

The Attempt at a Solution



The original range of inverse tan is ( -pi/2, pi/2)

Should tan inverse (2x) have (-pi, pi) as its range?
 
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No. 2x changes x values, the domain, not the range. The range of tan-1(x) is (-pi/2, pi/2) because as x goes to negative infinity, tan-1[/sup](x) goes to -pi/2 and as x goes to infinity, tan-1(x) goes to pi/2.

As x goes to -infinity, what does 2x go to? What does tan-1
(2x) go to?
As x goes to infinity, what does 2x go to? What does tan-1
(2x) go to?
 
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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