int frac{dx}{(1 + x^2)sqrt{1 - x^2}} = | vedclass

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(B) Let $I = \int \frac{dx}{(1 + x^2)\sqrt{1 - x^2}}$.Put $x = \sin 	heta$,then $dx = \cos 	heta \, d	heta$.Substituting these into the integral:$I

Vedclass · Accepted Answer

$\frac{1}{\sqrt{2}} 	an^{-1} \left[ \frac{x\sqrt{2}}{\sqrt{1 - x^2}} ight] + c$

Vedclass · Answer

(B) Let $I = \int \frac{dx}{(1 + x^2)\sqrt{1 - x^2}}$.Put $x = \sin 	heta$,then $dx = \cos 	heta \, d	heta$.Substituting these into the integral:$I = \int \frac{\cos 	heta \, d	heta}{(1 + \sin^2 	heta)\sqrt{1 - \sin^2 	heta}} = \int \frac{\cos 	heta \, d	heta}{(1 + \sin^2 	heta)\cos 	heta} = \int \frac{d	heta}{1 + \sin^2 	heta}$.Divide numerator and denominator by $\cos^2 	heta$:$I = \int \frac{\sec^2 	heta \, d	heta}{\sec^2 	heta + 	an^2 	heta} = \int \frac{\sec^2 	heta \, d	heta}{1 + 	an^2 	heta + 	an^2 	heta} = \int \frac{\sec^2 	heta \, d	heta}{1 + 2	an^2 	heta}$.Now,put $t = 	an 	heta$,so $dt = \sec^2 	heta \, d	heta$:$I = \int \frac{dt}{1 + 2t^2} = \frac{1}{2} \int \frac{dt}{t^2 + (1/\sqrt{2})^2}$.Using the formula $\int \frac{dx}{x^2 + a^2} = \frac{1}{a} 	an^{-1}(\frac{x}{a}) + c$:$I = \frac{1}{2} \cdot \frac{1}{1/\sqrt{2}} 	an^{-1} \left( \frac{t}{1/\sqrt{2}} ight) + c = \frac{1}{\sqrt{2}} 	an^{-1}(\sqrt{2}t) + c$.Since $t = 	an 	heta$ and $x = \sin 	heta$,we have $	an 	heta = \frac{\sin 	heta}{\cos 	heta} = \frac{x}{\sqrt{1 - x^2}}$.Therefore,$I = \frac{1}{\sqrt{2}} 	an^{-1} \left( \frac{x\sqrt{2}}{\sqrt{1 - x^2}} ight) + c$.

$\int \frac{dx}{(1 + x^2)\sqrt{1 - x^2}} = $

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