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

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(A) Let $I = \int x \operatorname{Cos}^{-1}\left(\frac{1-x^2}{1+x^2}ight) dx$. Substitute $x = \operatorname{Tan} 	heta$,so $dx = \operatorname{Sec

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$-x + (1+x^2) \operatorname{Tan}^{-1} x + c$

Vedclass · Answer

(A) Let $I = \int x \operatorname{Cos}^{-1}\left(\frac{1-x^2}{1+x^2}ight) dx$. Substitute $x = \operatorname{Tan} 	heta$,so $dx = \operatorname{Sec}^2 	heta d	heta$. Since $x > 0$,$	heta \in (0, \pi/2)$. The expression becomes $\operatorname{Cos}^{-1}\left(\frac{1-\operatorname{Tan}^2 	heta}{1+\operatorname{Tan}^2 	heta}ight) = \operatorname{Cos}^{-1}(\operatorname{Cos} 2	heta) = 2	heta$. Thus,$I = \int \operatorname{Tan} 	heta (2	heta) \operatorname{Sec}^2 	heta d	heta = 2 \int 	heta \operatorname{Tan} 	heta \operatorname{Sec}^2 	heta d	heta$. Using integration by parts,let $u = 	heta$ and $dv = \operatorname{Tan} 	heta \operatorname{Sec}^2 	heta d	heta$. Then $du = d	heta$ and $v = \frac{\operatorname{Tan}^2 	heta}{2}$. $I = 2 \left[ 	heta \cdot \frac{\operatorname{Tan}^2 	heta}{2} - \int \frac{\operatorname{Tan}^2 	heta}{2} d	heta ight] = 	heta \operatorname{Tan}^2 	heta - \int (\operatorname{Sec}^2 	heta - 1) d	heta$. $I = 	heta \operatorname{Tan}^2 	heta - \operatorname{Tan} 	heta + 	heta + c = 	heta (1 + \operatorname{Tan}^2 	heta) - \operatorname{Tan} 	heta + c$. Substituting back $	heta = \operatorname{Tan}^{-1} x$ and $\operatorname{Tan} 	heta = x$: $I = (1+x^2) \operatorname{Tan}^{-1} x - x + c$.

$\int x \operatorname{Cos}^{-1}\left(\frac{1-x^2}{1+x^2}\right) d x$ for $x > 0$ is equal to:

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