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anonymous 5 years ago evaluate the integral (32x^3dx)/sqrt(10-x^4), using u = 10 - x^4

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1. anonymous

$=32 \int\limits \frac{x^3}{\sqrt(10-x^4)} dx$ let : $u = 10-x^4$ $du = -4x^3 dx$ so : $= 32(\frac{-1}{4}) \int\limits \frac{1}{\sqrt(u)} du$ $= -8 \int\limits u^{\frac{-1}{2}} du$ $= -8 [ 2\sqrt(u)] + c = -16 \sqrt(10-x^4) + c$ Hope it's clear now ^_^

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