int {xsqrt {frac{{1 - {x^2}}}{{1 + {x^2}}}} } | vedclass

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Question

(A) Let $I = \int x\sqrt {\frac{{1 - {x^2}}}{{1 + {x^2}}}} \;dx$. Multiply the numerator and denominator by $\sqrt{1-x^2}$: $I = \int \frac{x(1-x^2)}{

Vedclass · Accepted Answer

$\frac{1}{2}[{\sin ^{ - 1}}({x^2}) + \sqrt {1 - {x^4}} ] + c$

Vedclass · Answer

(A) Let $I = \int x\sqrt {\frac{{1 - {x^2}}}{{1 + {x^2}}}} \;dx$. Multiply the numerator and denominator by $\sqrt{1-x^2}$: $I = \int \frac{x(1-x^2)}{\sqrt{(1-x^2)(1+x^2)}} \;dx = \int \frac{x - x^3}{\sqrt{1-x^4}} \;dx$. Split the integral: $I = \int \frac{x}{\sqrt{1-x^4}} \;dx - \int \frac{x^3}{\sqrt{1-x^4}} \;dx$. For the first part,let $x^2 = t$,then $2x \;dx = dt$,so $x \;dx = \frac{1}{2} dt$: $\int \frac{x}{\sqrt{1-x^4}} \;dx = \frac{1}{2} \int \frac{dt}{\sqrt{1-t^2}} = \frac{1}{2} \sin^{-1}(t) = \frac{1}{2} \sin^{-1}(x^2)$. For the second part,let $1-x^4 = u$,then $-4x^3 \;dx = du$,so $x^3 \;dx = -\frac{1}{4} du$: $\int \frac{x^3}{\sqrt{1-x^4}} \;dx = -\frac{1}{4} \int u^{-1/2} \;du = -\frac{1}{4} (2\sqrt{u}) = -\frac{1}{2} \sqrt{1-x^4}$. Combining these results: $I = \frac{1}{2} \sin^{-1}(x^2) - (-\frac{1}{2} \sqrt{1-x^4}) + c = \frac{1}{2} [\sin^{-1}(x^2) + \sqrt{1-x^4}] + c$.

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

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