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

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(A) Let $I = \int \frac{1 + x^2}{\sqrt{1 - x^2}} dx$.Substitute $x = \sin 	heta$,so $dx = \cos 	heta d	heta$.The integral becomes $I = \int \frac{1

Vedclass · Accepted Answer

$\frac{3}{2} \sin^{-1} x - \frac{1}{2} x \sqrt{1 - x^2} + c$

Vedclass · Answer

(A) Let $I = \int \frac{1 + x^2}{\sqrt{1 - x^2}} dx$.Substitute $x = \sin 	heta$,so $dx = \cos 	heta d	heta$.The integral becomes $I = \int \frac{1 + \sin^2 	heta}{\cos 	heta} \cos 	heta d	heta = \int (1 + \sin^2 	heta) d	heta$.Using the identity $\sin^2 	heta = \frac{1 - \cos 2	heta}{2}$,we get $I = \int (1 + \frac{1 - \cos 2	heta}{2}) d	heta = \int (\frac{3}{2} - \frac{1}{2} \cos 2	heta) d	heta$.Integrating,we get $I = \frac{3}{2} 	heta - \frac{1}{4} \sin 2	heta + c = \frac{3}{2} 	heta - \frac{1}{2} \sin 	heta \cos 	heta + c$.Since $	heta = \sin^{-1} x$ and $\cos 	heta = \sqrt{1 - \sin^2 	heta} = \sqrt{1 - x^2}$,we have $I = \frac{3}{2} \sin^{-1} x - \frac{1}{2} x \sqrt{1 - x^2} + c$.

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

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