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1.Show that for all $a > 0$

$$ \int_{0}^{\pi} (1 + a\sin^2(x))^{-1} \mathrm{d}x = 2 \int_{0}^{\frac{\pi}{2}} (1 + a\sin^2(x))^{-1}\mathrm{d}x $$

  1. Show that for all $ n \in \mathbb{N}^* $

$$ \int_{0}^{n\pi} (1 + x^3\sin^2(x))^{-1} \mathrm{d}x = \sum_{k=0}^{n-1} \int_{0}^{\pi} (1 + (k\pi + x)^3\sin^2(x))^{-1}\mathrm{d}x $$

  1. Show that $$ \int_{0}^{\infty} (1 + x^3\sin^2(x))^{-1}\mathrm{d}x$$ converges.
  1. $$ \int_{0}^{\pi} (1 + a\sin^2(x))^{-1} \mathrm{d}x = \int_{0}^{\frac{\pi}{2}} (1 + a\sin^2(x))^{-1} \mathrm{d}x + \int_{\frac{\pi}{2}}^{\pi} (1 + a\sin^2(x))^{-1} \mathrm{d}x $$

I don't see how to show that $\int_{0}^{\pi} (1 + a\sin^2(x))^{-1} \mathrm{d}x = \int_{\frac{\pi}{2}}^{\pi} (1 + a\sin^2(x))^{-1}\mathrm{d}x $

  1. I'm stuck on this one.

  2. I need to show that $ \int_{0}^{\pi} (1 + (k\pi + x)^3\sin^2(x))^{-1}\mathrm{d}x $ converges, I don't know how. Thank you for your help.

Masacroso
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Zouhair El Yaagoubi
  • 1,193

1 Answers1

1

  1. Try the change of variables $u=\pi-x$.

  2. Properly cut the integral into $n$ parts and use a change of variables on each part.

  3. We need to control the generic term of the series

$$\int_0^\pi (1+(k\pi+x)^3 \sin^2(x))^{-1} \mathrm{d}x \leq \int_0^\pi (1+(k\pi)^3 \sin^2(x))^{-1} \mathrm{d}x. $$

Solution 1 : Use the change of variable $u=\tan{x}$ to show that $$\int_0^{\pi/2} (1+a\sin^2{x})^{-1} \mathrm{d}x=\frac{\pi}{2\sqrt{1+a}}.$$

Solution 2 : Use the inequality $\sin{x}\geq (2/\pi)x$ for $0\leq x \leq \pi/2$ to show that

$$\int_0^{\pi/2} (1+a\sin^2{x})^{-1} \mathrm{d}x \leq \frac{\pi^2}{4\sqrt{a}}.$$

tristan
  • 2,883