How do you integrate by substitution $\int \sqrt[3]{1 - 2 x^{2}} (- 4 x) d x$ $int root3(1-2x^2)(-4x)dx$ ?

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Answer 1 · 2016-10-25T11:22:47

$= \frac{3}{4} {(1 - 2 x^{2})}^{\frac{4}{3}} + C$ $= 3/4 (1- 2x^2)^(4/3) + C$

let $y = 1 - 2 x^{2}, d y = (- 4 x) d x$ $y = 1- 2x^2, dy = color(red)((- 4x) dx)$

$\int \sqrt[3]{1 - 2 x^{2}} (- 4 x) d x$ $int root3(1-2x^2)color(red)((-4x) dx)$

$= int root3y \ color(red)(dy)$

$= int y^(1/3) \ color(red)(dy)$

$= (3 y^(4/3))/4 + C$

$= 3/4 (1- 2x^2)^(4/3) + C$

How do you integrate by substitution int root3(1-2x^2)(-4x)dx∫3√1−2x2(−4x)dxint root3(1-2x^2)(-4x)dx?