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Question Number 6439 by sanusihammed last updated on 27/Jun/16

$$H={\int }_0^{\frac{\mathrm{\Pi }}{4}}\sqrt{tanx}dx$$

Commented by Temp last updated on 27/Jun/16

$$\int \sqrt{\tan x}dx\mathrm{is}\mathrm{difficult}\mathrm{to}\mathrm{solve}.$$

Commented by nburiburu last updated on 27/Jun/16

$$bysubstitution:t=\sqrt{tanx}\Rightarrow t^2=tanx$$$$2tdt={sec}^{2}xdx=(1+{tan}^{2}x)dx$$$$dx=\frac{2t}{1+t^4}dt$$$$\int t.\frac{2t}{1+t^4}dt=\int \frac{2t^2}{1+t^4}dt$$$$andfromhereitisrarebutsimplierwithcomplexrootsinarationaldescomposition.$$

Commented by prakash jain last updated on 27/Jun/16

$$\mathrm{Question}119\mathrm{has}\mathrm{answer}\mathrm{for}\int \sqrt{\tan \theta }$$$$\frac{1}{\sqrt{2}}{\tan }^{-1}\left(\frac{\tan \theta -1}{\sqrt{2\tan \theta }}\right)+\frac{1}{2\sqrt{2}}\ln \mid \frac{\tan \theta +1-\sqrt{2\tan \theta }}{\tan \theta +1+\sqrt{2\tan \theta }}\mid +C$$

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