int_{frac{1}{2}}^2 frac{1}{x} operatorname{co | vedclass

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(A) Let $I = \int_{\frac{1}{2}}^2 \frac{1}{x} \operatorname{cosec}^{101}\left(x-\frac{1}{x}\right) d x$. Using the property $\int_a^b f(x) dx = \int_a

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(A) Let $I = \int_{\frac{1}{2}}^2 \frac{1}{x} \operatorname{cosec}^{101}\left(x-\frac{1}{x}ight) d x$. Using the property $\int_a^b f(x) dx = \int_a^b f\left(\frac{ab}{x}ight) \frac{ab}{x^2} dx$,we set $a = \frac{1}{2}$ and $b = 2$,so $ab = 1$. Then $I = \int_{\frac{1}{2}}^2 \frac{1}{1/x} \operatorname{cosec}^{101}\left(\frac{1}{x}-xight) \frac{1}{x^2} dx$. $I = \int_{\frac{1}{2}}^2 x \operatorname{cosec}^{101}\left(-(x-\frac{1}{x})ight) \frac{1}{x^2} dx$. Since $\operatorname{cosec}(- 	heta) = -\operatorname{cosec}(	heta)$ and the power $101$ is odd,$\operatorname{cosec}^{101}(- 	heta) = -\operatorname{cosec}^{101}(	heta)$. Thus,$I = - \int_{\frac{1}{2}}^2 \frac{1}{x} \operatorname{cosec}^{101}\left(x-\frac{1}{x}ight) dx = -I$. $2I = 0 \implies I = 0$.

$\int_{\frac{1}{2}}^2 \frac{1}{x} \operatorname{cosec}^{101}\left(x-\frac{1}{x}\right) d x=$

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