int frac{x^3}{sqrt{1 - x^8}} , dx = | vedclass

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Question

(C) Let $I = \int \frac{x^3}{\sqrt{1 - x^8}} \, dx$. Rewrite the integral as $I = \int \frac{x^3}{\sqrt{1 - (x^4)^2}} \, dx$. Substitute $t = x^4$,the

Vedclass · Accepted Answer

$\frac{1}{4} \sin^{-1}(x^4) + c$

Vedclass · Answer

(C) Let $I = \int \frac{x^3}{\sqrt{1 - x^8}} \, dx$. Rewrite the integral as $I = \int \frac{x^3}{\sqrt{1 - (x^4)^2}} \, dx$. Substitute $t = x^4$,then $dt = 4x^3 \, dx$,which implies $x^3 \, dx = \frac{1}{4} \, dt$. Substituting these into the integral,we get $I = \int \frac{1}{\sqrt{1 - t^2}} \cdot \frac{1}{4} \, dt$. $I = \frac{1}{4} \int \frac{1}{\sqrt{1 - t^2}} \, dt$. Using the standard integral $\int \frac{1}{\sqrt{1 - t^2}} \, dt = \sin^{-1}(t) + c$,we get $I = \frac{1}{4} \sin^{-1}(t) + c$. Substituting back $t = x^4$,we obtain $I = \frac{1}{4} \sin^{-1}(x^4) + c$.

$\int \frac{x^3}{\sqrt{1 - x^8}} \, dx = $

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