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Question Number 130809 by bemath last updated on 29/Jan/21

$$\int \tan \mathrm{x}\sqrt{\sin \mathrm{x}}\mathrm{dx}?$$

Answered by EDWIN88 last updated on 29/Jan/21

$$let\sqrt{\sin x}=u\Rightarrow \sin x=u^2$$$$\tan x=\frac{u^2}{\sqrt{1-u^4}}$$$$dx=\frac{2udu}{\sqrt{1-u^4}}$$$$I=\int \frac{u^3}{\sqrt{1-u^4}}\left(\frac{2u}{\sqrt{1-u^4}}\right)du$$$$I=\int \frac{2u^4}{1-u^4}du=\int (-2+\frac{2}{1-u^4})du$$$$I=-2u+2\int \frac{du}{(1+u^2)(1-u^2)}$$$$I=-2u+\int (\frac{1}{1-u^2}+\frac{1}{u^2+1})du$$$$I=-2u+{\tan }^{-1}\left(u\right)+\frac{1}{2}\int (\frac{1}{1-u}+\frac{1}{1+u})du$$$$I=-2u+{\tan }^{-1}\left(u\right)+\frac{1}{2}\ln \mid 1-u^2\mid +c$$$$I=-2\sqrt{\sin x}+{\tan }^{-1}\left(\sqrt{\sin x}\right)+\frac{1}{2}\ln \mid 1-\sin x\mid +c$$

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