y=int cos left{2 tan ^{-1} sqrt{frac{1-x}{1+x | vedclass

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(D) Let $I = \int \cos \left(2 	an ^{-1} \sqrt{\frac{1-x}{1+x}}ight) dx$. Substitute $x = \cos 2	heta$, so $dx = -2 \sin 2	heta d	heta$. Then, $

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parabolas

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(D) Let $I = \int \cos \left(2 	an ^{-1} \sqrt{\frac{1-x}{1+x}}ight) dx$. Substitute $x = \cos 2	heta$, so $dx = -2 \sin 2	heta d	heta$. Then, $I = \int \cos \left(2 	an ^{-1} \sqrt{\frac{1-\cos 2	heta}{1+\cos 2	heta}}ight) (-2 \sin 2	heta) d	heta$. Using the identity $\frac{1-\cos 2	heta}{1+\cos 2	heta} = 	an^2 	heta$, we get $I = -2 \int \cos(2 	an^{-1}(	an 	heta)) \sin 2	heta d	heta$. This simplifies to $I = -2 \int \cos(2	heta) \sin 2	heta d	heta$. Using the identity $\sin 4	heta = 2 \sin 2	heta \cos 2	heta$, we have $I = -\int \sin 4	heta d	heta = \frac{\cos 4	heta}{4} + C$. Since $\cos 2	heta = x$, then $\cos 4	heta = 2 \cos^2 2	heta - 1 = 2x^2 - 1$. Thus, $y = \frac{2x^2 - 1}{4} + C = \frac{1}{2}x^2 + C'$, which represents a family of parabolas.

$y=\int \cos \left\{2 \tan ^{-1} \sqrt{\frac{1-x}{1+x}}\right\} d x$ is an equation of a family of

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