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$\large\int (sin \theta)^{\dfrac{1}{2}} d\theta$

$=\large \int \sqrt{sin \theta} d\theta$

$= \large \int \sqrt{2 cos^2 \left(\dfrac{\theta}{2}- \dfrac{\pi}{4}\right) - 1} d\theta $ $\qquad \text{Using Trigonometric Identities}$

$= \large \int \sqrt{1-2sin^2 \left(\dfrac{\theta}{2}- \dfrac{\pi}{4} \right)} d\theta$

$= \large \int \sqrt{1-2sin^2\left(\dfrac{2\theta - \pi}{4}\right )} d\theta$

Now, assuming

$\qquad \qquad \dfrac{2\theta - \pi}{4} = u$

$\qquad \qquad Or, \dfrac{d}{d\theta}\left(\dfrac{2\theta - \pi}{4}\right) = du$

$\qquad \qquad Or, \dfrac{1}{4}\{\dfrac{d}{d\theta}(2x)+\dfrac{d}{d\theta}(-\pi)\} = du$

$\qquad \qquad Or, \dfrac{2}{4}d\theta=du$

$\qquad\qquad Or, d\theta = 2du$

∴ $\int \sqrt{1-2sin^2\left(\dfrac{2\theta - \pi}{4}\right)}d\theta= \large 2 \int \sqrt{1-2sin^2(u)}du$

This structure $\int \sqrt{1-2sin^2(u)}du$ is a special integral ( incomplete elliptic integral of the second kind )

∴ $\large \int \sqrt{1-2sin^2(u)}du = E(u \mid 2) \hspace{0.4cm} \rightarrow \text{Here E is Elliptic E}$

∴ $\large2\int \sqrt{1-2sin^2(u)} du = 2.E(u \mid 2)$

Now, putting back $\dfrac{2\theta - \pi}{4} = u$

∴ $\color{green}{2.E(u \mid 2) = 2.E\left(\dfrac{2\theta - \pi}{4} \mid 2\right) +c}$ where, c = constant
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