Question Number 140490 by benjo_mathlover last updated on 08/May/21
Commented by john_santu last updated on 08/May/21
$$\frac{\pi}{\mathrm{2}}\:\left(\sqrt{\mathrm{2}\sqrt{\mathrm{2}}+\mathrm{2}}\:−\mathrm{2}\:\right)\:\approx\:\mathrm{0}.\mathrm{31} \\ $$
Answered by MJS_new last updated on 08/May/21
![∫(√(tan x (1−tan x)))dx= [t=((√(tan x))/( (√(1−tan x)))) → dx=2cos^2 x (√(tan x (1−tan x)^3 ))dt] =∫(t^2 /((t^2 +1)(t^4 +t^2 +(1/2))))dt now decompose and solve. I′ll complete it later](Q140496.png)
$$\int\sqrt{\mathrm{tan}\:{x}\:\left(\mathrm{1}−\mathrm{tan}\:{x}\right)}{dx}= \\ $$$$\:\:\:\:\:\left[{t}=\frac{\sqrt{\mathrm{tan}\:{x}}}{\:\sqrt{\mathrm{1}−\mathrm{tan}\:{x}}}\:\rightarrow\:{dx}=\mathrm{2cos}^{\mathrm{2}} \:{x}\:\sqrt{\mathrm{tan}\:{x}\:\left(\mathrm{1}−\mathrm{tan}\:{x}\right)^{\mathrm{3}} }{dt}\right] \\ $$$$=\int\frac{{t}^{\mathrm{2}} }{\left({t}^{\mathrm{2}} +\mathrm{1}\right)\left({t}^{\mathrm{4}} +{t}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}\right)}{dt} \\ $$$$\mathrm{now}\:\mathrm{decompose}\:\mathrm{and}\:\mathrm{solve}.\:\mathrm{I}'\mathrm{ll}\:\mathrm{complete}\:\mathrm{it}\:\mathrm{later} \\ $$