\( \int_{1/\sqrt{3}}^{\sqrt{3}}\frac{dx}{\left ( x^3+1 \right )\left ( x^2+1 \right )} \)
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\( \int_{1/\sqrt{3}}^{\sqrt{3}}\frac{dx}{\left ( x^3+1 \right )\left ( x^2+1 \right )} \)
Prove that: $$\int_{1/\sqrt{3}}^{\sqrt{3}}\frac{dx}{\left ( x^3+1 \right )\left ( x^2+1 \right )}=\frac{\pi}{12}$$
Imagination is much more important than knowledge.
Re: \( \int_{1/\sqrt{3}}^{\sqrt{3}}\frac{dx}{\left ( x^3+1 \right )\left ( x^2+1 \right )} \)
Let \(\displaystyle I = \int_{\alpha^{-1}}^{\alpha}\frac{1}{(x^3+1)(x^2+1)}dx....................................................(1)\), Where \(\displaystyle \alpha = \sqrt{3}\)
Now Let \(\displaystyle x=\frac{1}{t}\) and \(\displaystyle dx = -\frac{1}{t^2}\) and changing Limits
\(\displaystyle I = \int_{\alpha}^{\alpha^{-1}}\frac{t^3\cdot t^2}{(1+t^3)(1+t^2)}\cdot -\frac{1}{t^2}dt = \int_{\alpha^{-1}}^{\alpha}\frac{t^3}{(1+t^3)(1+t^2)}dt\)
So We get \(\displaystyle I = \int_{\alpha^{-1}}^{\alpha}\frac{x^3}{(1+x^3)(1+x^2)}dx.............................................(2)\)
Now Add \(\displaystyle (1)\) and \(\displaystyle (2)\)
So \(\displaystyle 2I = \int_{\alpha^{-1}}^{\alpha}\frac{(x^3+1)}{(x^3+1)(x^2+1)}dx = \int_{\alpha^{-1}}^{\alpha}\frac{1}{1+x^2}dx\)
So \(\displaystyle 2I = \tan^{-1}\left(\alpha)\right)- \tan^{-1}\left(\frac{1}{\alpha}\right) = \frac{\pi}{3}-\frac{\pi}{6} = \frac{\pi}{6}\), Using \(\alpha = \sqrt{3}\)
So we get \(\displaystyle I = \int_{\frac{1}{\sqrt{3}}}^{\sqrt{3}}\frac{1}{(x^3+1)(x^2+1)}dx = \frac{\pi}{12}\)
Now Let \(\displaystyle x=\frac{1}{t}\) and \(\displaystyle dx = -\frac{1}{t^2}\) and changing Limits
\(\displaystyle I = \int_{\alpha}^{\alpha^{-1}}\frac{t^3\cdot t^2}{(1+t^3)(1+t^2)}\cdot -\frac{1}{t^2}dt = \int_{\alpha^{-1}}^{\alpha}\frac{t^3}{(1+t^3)(1+t^2)}dt\)
So We get \(\displaystyle I = \int_{\alpha^{-1}}^{\alpha}\frac{x^3}{(1+x^3)(1+x^2)}dx.............................................(2)\)
Now Add \(\displaystyle (1)\) and \(\displaystyle (2)\)
So \(\displaystyle 2I = \int_{\alpha^{-1}}^{\alpha}\frac{(x^3+1)}{(x^3+1)(x^2+1)}dx = \int_{\alpha^{-1}}^{\alpha}\frac{1}{1+x^2}dx\)
So \(\displaystyle 2I = \tan^{-1}\left(\alpha)\right)- \tan^{-1}\left(\frac{1}{\alpha}\right) = \frac{\pi}{3}-\frac{\pi}{6} = \frac{\pi}{6}\), Using \(\alpha = \sqrt{3}\)
So we get \(\displaystyle I = \int_{\frac{1}{\sqrt{3}}}^{\sqrt{3}}\frac{1}{(x^3+1)(x^2+1)}dx = \frac{\pi}{12}\)
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