# Integration of $\int\sqrt{\tan\theta} \cdot d\theta$ [duplicate]

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

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

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 $$1+u^4=(1-2\sqrt u+u^2)(1+2\sqrt u+u^2)$$