Evaluate the integral int x^2 sqrt{8-x^6} , d | vedclass

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Question

(B) Let $I = \int x^2 \sqrt{8-x^6} \, dx$. Substitute $u = x^3$,then $du = 3x^2 \, dx$,which implies $x^2 \, dx = \frac{1}{3} du$. The integral become

Vedclass · Accepted Answer

$\frac{1}{3}\left[\frac{x^3}{2} \sqrt{8-x^6}+4 \sin ^{-1} \frac{x^3}{2 \sqrt{2}}\right]$

Vedclass · Answer

(B) Let $I = \int x^2 \sqrt{8-x^6} \, dx$. Substitute $u = x^3$,then $du = 3x^2 \, dx$,which implies $x^2 \, dx = \frac{1}{3} du$. The integral becomes $I = \frac{1}{3} \int \sqrt{8-u^2} \, du$. Using the standard formula $\int \sqrt{a^2-u^2} \, du = \frac{u}{2} \sqrt{a^2-u^2} + \frac{a^2}{2} \sin^{-1} \left(\frac{u}{a}\right) + C$,where $a^2 = 8$ (so $a = 2\sqrt{2}$): $I = \frac{1}{3} \left[ \frac{u}{2} \sqrt{8-u^2} + \frac{8}{2} \sin^{-1} \left(\frac{u}{2\sqrt{2}}\right) \right] + C$. Substituting $u = x^3$ back: $I = \frac{1}{3} \left[ \frac{x^3}{2} \sqrt{8-x^6} + 4 \sin^{-1} \left(\frac{x^3}{2\sqrt{2}}\right) \right] + C$. Thus,the correct option is $B$.

Evaluate the integral $\int x^2 \sqrt{8-x^6} \, dx$. Given $|x| < \sqrt{2}$.

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