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Integral of 2x sqrt(1-x^2)
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let u = 1-x^2 so du = -2xdx substitute it into the formula and you get integral of -(u)^1/2du after everythign is done youre left with u^3/2 / (3/2) +c after u take everything out u'll get -2/3 * (1-x^2)^3/2 + c
\[\int\limits2x \sqrt{1-x^2}dx\] \[2\int\limits x \sqrt{1-x^2}dx\] u = 1-x^2 du = -2x dx \[-\int\limits \sqrt{u}du\] \[=\frac{2u^{\frac{3}{2}}}{3}+C\] \[= -\frac{2}{3}(1-x^2)^{\frac{3}{2}}\]
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