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According to wolfram, the following equation holds.
However, I do not understand the derivation process.
Could you please tell me how to derive it?
$$\int_0^1 \frac{u^4}{\sqrt{(1 - u^2) (1 - k u^2)}} du = \frac{(k + 2) K(k) - 2 (k + 1) E(k)}{3 k^2}\quad(0\leq k \leq 1)$$ where K and E are complete elliptic integrals of the first kind and complete elliptic integrals of the second kind.

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  • $\begingroup$ $$\frac{u^4}{\sqrt{1-u^2}\sqrt{1-ku^2}}=\frac{u^4}{1-ku^2}\cdot\frac{\sqrt{1-ku^2}}{\sqrt{1-u^2}}$$Expand into partial fractions and use an [earlier result](math.stackexchange.com/questions/4693466/…) to get$$k^2I=K(k)-E(k)-\int_0^1\frac{ku^2\sqrt{1-ku^2}}{\sqrt{1-u^2}}\,du$$but I'm not sure how to proceed just yet... $\endgroup$
    – user170231
    May 7, 2023 at 6:16

1 Answer 1

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Hint

$$I=\int \frac{u^4}{\sqrt{(1 - u^2) (1 - k u^2)}} du $$

$$u=\sin(x) \qquad \implies \qquad I=\int \frac{\sin ^4(x)}{\sqrt{1-k \sin ^2(x)}}\, dx$$ Using $$\sin^4(x)=(\sin^2(x)+1)(\sin^2(x)-1)+1$$ the problem is quite simple.

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