Question

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

Answer 1

\[=32 \int\limits \frac{x^3}{\sqrt(10-x^4)} dx\]

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 ^_^