int left( frac{tan left( frac{1}{x} right)}{x | vedclass

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Question

(B) Let $I = \int \left( \frac{	an \left( \frac{1}{x} ight)}{x} ight)^2 \, dx$. Substitute $t = \frac{1}{x}$,then $dt = -\frac{1}{x^2} \, dx$,whi

Vedclass · Accepted Answer

$\frac{1}{x} - 	an \left( \frac{1}{x} ight) + c$,where $c$ is a constant of integration.

Vedclass · Answer

(B) Let $I = \int \left( \frac{	an \left( \frac{1}{x} ight)}{x} ight)^2 \, dx$. Substitute $t = \frac{1}{x}$,then $dt = -\frac{1}{x^2} \, dx$,which implies $dx = -x^2 \, dt = -\frac{1}{t^2} \, dt$. Substituting these into the integral: $I = \int 	an^2(t) \cdot \frac{1}{x^2} \cdot (-x^2 \, dt) = -\int 	an^2(t) \, dt$. Using the identity $	an^2(t) = \sec^2(t) - 1$: $I = -\int (\sec^2(t) - 1) \, dt = \int (1 - \sec^2(t)) \, dt$. Integrating term by term: $I = t - 	an(t) + c$. Substituting back $t = \frac{1}{x}$: $I = \frac{1}{x} - 	an \left( \frac{1}{x} ight) + c$.

$\int \left( \frac{\tan \left( \frac{1}{x} \right)}{x} \right)^2 \, dx =$

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