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Question Number 71695 by peter frank last updated on 18/Oct/19

Answered by MJS last updated on 19/Oct/19

$$\int \frac{\sqrt{\tan x}}{1+\sqrt{\tan x}}dx=$$$$[t=\sqrt{\tan x}\to dx=2\sqrt{\tan x}{\cos }^{2}xdt=\frac{2t}{t^4+1}dt]$$$$=2\int \frac{t^2}{(t+1)(t^4+1)}dt=$$$$=2\int \frac{t^2}{(t+1)(t^2-\sqrt{2}t+1)(t^2+\sqrt{2}t+1)}dt=$$$$=\int \frac{dt}{t+1}-\frac{1-\sqrt{2}}{2}\int \frac{t+1}{t^2-\sqrt{2}t+1}dt-\frac{1+\sqrt{2}}{2}\int \frac{t+1}{t^2+\sqrt{2}t+1}dt=$$$$=\ln (t+1)-$$$$-\frac{1-\sqrt{2}}{4}\ln (t^2-\sqrt{2}t+1)+\frac{1}{2}\arctan (\sqrt{2}t-1)-y$$$$-\frac{1+\sqrt{2}}{4}\ln (t^2+\sqrt{2}t+1)-\frac{1}{2}\arctan (\sqrt{2}t+1)$$$$\mathrm{and}\mathrm{now}\mathrm{put}t=\sqrt{\tan x}$$

Commented by peter frank last updated on 19/Oct/19

$$thankyou$$

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