What is the antiderivative of $sqrt(sinx)$ ?

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Answer 1 · 2017-02-03T18:19:45

This is an advanced kind of antiderivative.

According to WolframAlpha, it is $-2E(1/4(pi - 2x)|2) +C$ . Where $E(x|m)$ is the elliptic integral of the second kind with parameter $m = k^2$ .

No. I can't explain that.

What is the antiderivative of sqrt(sinx)sqrt(sinx)?