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$\int_0^\pi \frac{\cos x}{\sqrt{1-\sin ^2 x}} d x=$

The true solution set of the inequality $\sqrt{5x-6-x^2} + \left( \frac{\pi}{2} \int_{0}^{x} dz \right) > x \int_{0}^{\pi} \sin^2 x dx$ is:

If $I_n = \int_0^{\pi/4} \tan^n \theta \, d\theta$ for $n = 1, 2, 3, \ldots$,then $I_{n-1} + I_{n+1}$ is equal to

$\int_0^{1/\sqrt{2}} \frac{\sin^{-1}x}{(1-x^2)^{3/2}} dx = $

$\int_0^1 {{e^{2\ln x}}dx} = $

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