int_0^1 frac{x^7}{sqrt{1 - x^4}} dx is equal | vedclass

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(B) Let $I = \int_0^1 \frac{x^7}{\sqrt{1 - x^4}} dx = \int_0^1 \frac{x^6 \cdot x}{\sqrt{1 - x^4}} dx$.Substitute $t = x^4$,so $dt = 4x^3 dx$. This app

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$\frac{1}{3}$

Vedclass · Answer

(B) Let $I = \int_0^1 \frac{x^7}{\sqrt{1 - x^4}} dx = \int_0^1 \frac{x^6 \cdot x}{\sqrt{1 - x^4}} dx$.Substitute $t = x^4$,so $dt = 4x^3 dx$. This approach is slightly complex,so let's use $u = x^4$,$du = 4x^3 dx$. Then $x^4 = u$ and $x^2 = \sqrt{u}$.$I = \int_0^1 \frac{x^4 \cdot x^3}{\sqrt{1 - x^4}} dx = \int_0^1 \frac{u}{\sqrt{1 - u}} \cdot \frac{1}{4} du$.Let $1 - u = v^2$,then $du = -2v dv$. When $u=0, v=1$; when $u=1, v=0$.$I = \frac{1}{4} \int_1^0 \frac{1 - v^2}{v} (-2v) dv = \frac{1}{2} \int_0^1 (1 - v^2) dv$.$I = \frac{1}{2} [v - \frac{v^3}{3}]_0^1 = \frac{1}{2} (1 - \frac{1}{3}) = \frac{1}{2} \cdot \frac{2}{3} = \frac{1}{3}$.

$\int_0^1 \frac{x^7}{\sqrt{1 - x^4}} dx$ is equal to

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