Integrate the function: frac{x^{3}}{sqrt{1-x^ | vedclass

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(A) To integrate $\int \frac{x^{3}}{\sqrt{1-x^{8}}} dx$,we use the method of substitution.Let $t = x^{4}$.Then,differentiating both sides with respect

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) To integrate $\int \frac{x^{3}}{\sqrt{1-x^{8}}} dx$,we use the method of substitution.Let $t = x^{4}$.Then,differentiating both sides with respect to $x$,we get $\frac{dt}{dx} = 4x^{3}$,which implies $x^{3} dx = \frac{1}{4} dt$.Substituting these into the integral:$\int \frac{x^{3}}{\sqrt{1-(x^{4})^{2}}} dx = \int \frac{1}{\sqrt{1-t^{2}}} \cdot \frac{1}{4} dt$$= \frac{1}{4} \int \frac{1}{\sqrt{1-t^{2}}} dt$Using the standard integral formula $\int \frac{1}{\sqrt{1-t^{2}}} dt = \sin^{-1}(t) + C$,we get:$= \frac{1}{4} \sin^{-1}(t) + C$Substituting $t = x^{4}$ back into the expression:$= \frac{1}{4} \sin^{-1}(x^{4}) + C$

Integrate the function: $\frac{x^{3}}{\sqrt{1-x^{8}}}$

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