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Integration of $\int\sqrt{\tan\theta} \cdot d\theta$ is what.

I have tried to substitute $\tan\theta$ as $t$ but to no avail.

I do not know how to proceed. Please help.

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Hint

Define $t=\tan\theta$ and $u=\sqrt t$ $$\int\sqrt{\tan \theta}d\theta{=\int{\sqrt{\tan \theta}\over 1+\tan^2 \theta}(1+\tan^2 \theta)d\theta\\=\int{\sqrt t\over 1+t^2}dt\\=\int {2u^2\over 1+u^4}du\\}$$and expand $$1+u^4=(1-2\sqrt u+u^2)(1+2\sqrt u+u^2)$$

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