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一个简单的椭圆积分

PS 感谢 Fran 佬指出疏忽

∫1sin⁡xdx=∫1cos⁡(x−π2)dx=∫11−2sin⁡2(x2−π4)dx=2∫11−2sin⁡2(x2−π4)d(x2−π4)=2F(x2−π4∣2)\begin{aligned} \int \frac{1}{\sqrt{\sin x}}\mathrm{d}x&=\int \frac{1}{\sqrt{\cos\left( x-\frac{\pi}{2} \right)}}\mathrm{d}x\\ &=\int \frac{1}{\sqrt{1-2\sin^2\left( \frac{x}{2}-\frac{\pi}{4} \right)}}\mathrm{d}x\\ &=2\int \frac{1}{\sqrt{1-2\sin^2\left( \frac{x}{2}-\frac{\pi}{4} \right)}}\mathrm{d}\left( \frac{x}{2}-\frac{\pi}{4} \right)\\ &=2F\left(\left. \frac{x}{2}-\frac{\pi}{4}\right|2 \right) \end{aligned}