int frac{sqrt{x^2-a^2}}{x} d x = ____ | vedclass

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(A) Let $I = \int \frac{\sqrt{x^2-a^2}}{x} d x$. Substitute $x = a \sec 	heta$,then $d x = a \sec 	heta 	an 	heta d 	heta$. Substituting these in

Vedclass · Accepted Answer

$\sqrt{x^2-a^2}-a \sec ^{-1}\left(\frac{x}{a}\right)+c$

Vedclass · Answer

(A) Let $I = \int \frac{\sqrt{x^2-a^2}}{x} d x$. Substitute $x = a \sec 	heta$,then $d x = a \sec 	heta 	an 	heta d 	heta$. Substituting these into the integral: $I = \int \frac{\sqrt{a^2 \sec^2 	heta - a^2}}{a \sec 	heta} (a \sec 	heta 	an 	heta) d 	heta$ $I = \int \frac{a 	an 	heta}{a \sec 	heta} (a \sec 	heta 	an 	heta) d 	heta$ $I = a \int 	an^2 	heta d 	heta$ $I = a \int (\sec^2 	heta - 1) d 	heta$ $I = a (	an 	heta - 	heta) + c$ Since $x = a \sec 	heta$,we have $\sec 	heta = \frac{x}{a}$,so $	heta = \sec^{-1}(\frac{x}{a})$ and $	an 	heta = \sqrt{\sec^2 	heta - 1} = \sqrt{(\frac{x}{a})^2 - 1} = \frac{\sqrt{x^2-a^2}}{a}$. Substituting back: $I = a (\frac{\sqrt{x^2-a^2}}{a} - \sec^{-1}(\frac{x}{a})) + c$ $I = \sqrt{x^2-a^2} - a \sec^{-1}(\frac{x}{a}) + c$.

$\int \frac{\sqrt{x^2-a^2}}{x} d x = \_\_\_\_$

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