The integral frac{24}{pi} int_{0}^{sqrt{2}} f | vedclass

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(D) Let $I = \frac{24}{\pi} \int_{0}^{\sqrt{2}} \frac{(2-x^{2})}{(x^{2}+2) \sqrt{4+x^{4}}} dx$.Divide the numerator and denominator by $x^{2}$:$I = \f

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$3$

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(D) Let $I = \frac{24}{\pi} \int_{0}^{\sqrt{2}} \frac{(2-x^{2})}{(x^{2}+2) \sqrt{4+x^{4}}} dx$.Divide the numerator and denominator by $x^{2}$:$I = \frac{24}{\pi} \int_{0}^{\sqrt{2}} \frac{(\frac{2}{x^{2}}-1)}{(1+\frac{2}{x^{2}}) \sqrt{\frac{4}{x^{2}}+x^{2}}} dx$.Note that $\sqrt{4+x^{4}} = x \sqrt{x^{2}+\frac{4}{x^{2}}}$.$I = \frac{24}{\pi} \int_{0}^{\sqrt{2}} \frac{(\frac{2}{x^{2}}-1)}{(x+\frac{2}{x}) \sqrt{(x+\frac{2}{x})^{2}-4}} dx$.Let $t = x+\frac{2}{x}$,then $dt = (1-\frac{2}{x^{2}}) dx$.As $x 	o 0^{+}$,$t 	o \infty$. As $x 	o \sqrt{2}$,$t 	o \sqrt{2}+\frac{2}{\sqrt{2}} = 2\sqrt{2}$.$I = -\frac{24}{\pi} \int_{\infty}^{2\sqrt{2}} \frac{dt}{t \sqrt{t^{2}-4}} = \frac{24}{\pi} \int_{2\sqrt{2}}^{\infty} \frac{dt}{t \sqrt{t^{2}-4}}$.Using $\int \frac{dt}{t \sqrt{t^{2}-a^{2}}} = \frac{1}{a} \sec^{-1}(\frac{t}{a})$:$I = \frac{24}{\pi} [\frac{1}{2} \sec^{-1}(\frac{t}{2})]_{2\sqrt{2}}^{\infty} = \frac{12}{\pi} [\sec^{-1}(\infty) - \sec^{-1}(\sqrt{2})]$.$I = \frac{12}{\pi} [\frac{\pi}{2} - \frac{\pi}{4}] = \frac{12}{\pi} 	imes \frac{\pi}{4} = 3$.

The integral $\frac{24}{\pi} \int_{0}^{\sqrt{2}} \frac{(2-x^{2}) dx}{(2+x^{2}) \sqrt{4+x^{4}}}$ is equal to

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