The integral int {x,{{cos }^{ - 1}},left( {fr | vedclass

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(A) Let $I = \int x \cos^{-1} \left( \frac{1-x^2}{1+x^2} ight) dx$.Since $x > 0$,we use the substitution $x = 	an 	heta$,which implies $	heta = \

Vedclass · Accepted Answer

$- x + ( 1 + x^2)\, 	an^{-1} \,x + c$

Vedclass · Answer

(A) Let $I = \int x \cos^{-1} \left( \frac{1-x^2}{1+x^2} ight) dx$.Since $x > 0$,we use the substitution $x = 	an 	heta$,which implies $	heta = 	an^{-1} x$.Then $\cos^{-1} \left( \frac{1-x^2}{1+x^2} ight) = \cos^{-1} (\cos 2	heta) = 2	heta = 2	an^{-1} x$.Thus,$I = \int x (2	an^{-1} x) dx = 2 \int x 	an^{-1} x dx$.Using integration by parts,let $u = 	an^{-1} x$ and $dv = x dx$. Then $du = \frac{1}{1+x^2} dx$ and $v = \frac{x^2}{2}$.$I = 2 \left[ \frac{x^2}{2} 	an^{-1} x - \int \frac{x^2}{2(1+x^2)} dx ight] + c$.$I = x^2 	an^{-1} x - \int \frac{x^2+1-1}{1+x^2} dx + c$.$I = x^2 	an^{-1} x - \int \left( 1 - \frac{1}{1+x^2} ight) dx + c$.$I = x^2 	an^{-1} x - x + 	an^{-1} x + c$.$I = (x^2 + 1) 	an^{-1} x - x + c$.

The integral $\int {x\,{{\cos }^{ - 1}}\,\left( {\frac{{1 - {x^2}}}{{1 + {x^2}}}} \right)dx} \,\left( {x > 0} \right)$ is equal to

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