int operatorname{Cos}^{-1}left(frac{1-x^2}{1+ | vedclass

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(A) Let $x = 	an 	heta$,then $dx = \sec^2 	heta d	heta$. Substituting this into the integral,we get: $\int \operatorname{Cos}^{-1}\left(\frac{1-

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$2\left[x \operatorname{Tan}^{-1} x-\log \sqrt{1+x^2}\right]+c$

Vedclass · Answer

(A) Let $x = 	an 	heta$,then $dx = \sec^2 	heta d	heta$. Substituting this into the integral,we get: $\int \operatorname{Cos}^{-1}\left(\frac{1-	an^2 	heta}{1+	an^2 	heta}ight) \sec^2 	heta d	heta = \int \operatorname{Cos}^{-1}(\cos 2	heta) \sec^2 	heta d	heta = \int 2	heta \sec^2 	heta d	heta$. Using integration by parts,let $u = 2	heta$ and $dv = \sec^2 	heta d	heta$. Then $du = 2 d	heta$ and $v = 	an 	heta$. $\int 2	heta \sec^2 	heta d	heta = 2	heta 	an 	heta - \int 2 	an 	heta d	heta$. $= 2	heta 	an 	heta - 2 \ln|\sec 	heta| + c$. Since $x = 	an 	heta$,then $	heta = 	an^{-1} x$ and $\sec 	heta = \sqrt{1+	an^2 	heta} = \sqrt{1+x^2}$. Substituting back,we get: $2x 	an^{-1} x - 2 \ln(\sqrt{1+x^2}) + c = 2x 	an^{-1} x - \ln(1+x^2) + c$. This can be written as $2[x 	an^{-1} x - \frac{1}{2} \ln(1+x^2)] + c = 2[x 	an^{-1} x - \ln \sqrt{1+x^2}] + c$.

$\int \operatorname{Cos}^{-1}\left(\frac{1-x^2}{1+x^2}\right) d x=$

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