Question Number 195255 by Spillover last updated on 28/Jul/23
$$\int\frac{{dx}}{\mathrm{cos}\:^{\mathrm{3}} {x}\sqrt{\mathrm{4sin}\:{x}\mathrm{cos}\:{x}}} \\ $$
Answered by Frix last updated on 28/Jul/23
![[Using t=(√(tan x))] =∫(t^4 +1)dt=...=((5+tan^2 x)/5)(√(tan x)) +C](https://www.tinkutara.com/question/Q195266.png)
$$\left[\mathrm{Using}\:{t}=\sqrt{\mathrm{tan}\:{x}}\right] \\ $$$$=\int\left({t}^{\mathrm{4}} +\mathrm{1}\right){dt}=…=\frac{\mathrm{5}+\mathrm{tan}^{\mathrm{2}} \:{x}}{\mathrm{5}}\sqrt{\mathrm{tan}\:{x}}\:+{C} \\ $$
Answered by Spillover last updated on 03/Aug/23
$$\int\frac{{dx}}{\mathrm{cos}\:^{\mathrm{3}} {x}\sqrt{\mathrm{4sin}\:{x}\mathrm{cos}\:{x}}} \\ $$$$=\int\frac{{dx}}{\mathrm{2cos}\:^{\mathrm{4}} {x}\sqrt{\mathrm{tan}\:{x}}}=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{dx}}{\mathrm{cos}\:^{\mathrm{4}} {x}\sqrt{\mathrm{tan}\:{x}}} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{dx}\left(\mathrm{1}+\mathrm{tan}\:^{\mathrm{2}} {x}\right)\mathrm{sec}\:^{\mathrm{2}} {x}}{\:\sqrt{\mathrm{tan}\:{x}}} \\ $$$${u}=\sqrt{\mathrm{tan}\:{x}}\:\:\:\:\:\:\Rightarrow\:\underset{{spillover}} {\:{u}^{\mathrm{2}} =\mathrm{tan}\:{x}}\:\:\:\:\Rightarrow\:\:\mathrm{2}{udu}=\mathrm{sec}\:^{\mathrm{2}} {x} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{u}^{\mathrm{4}} +\mathrm{1}}{{u}}.\mathrm{2}{u}\:\:\: \\ $$$$\:\int\left(\mathrm{1}+{u}^{\mathrm{4}} \right){du}\:\:\Rightarrow\:{u}+\frac{{u}^{\mathrm{5}} }{\mathrm{5}}+{D} \\ $$$$\sqrt{\mathrm{tan}\:{x}}+\frac{\left(\sqrt{\mathrm{tan}\:{x}}\:\right)^{\mathrm{5}} }{\mathrm{5}}+{D} \\ $$