Question Number 216990 by MathematicalUser2357 last updated on 26/Feb/25
$$\left(\mathrm{1}\right)\:\int\left(\mathrm{sec}^{\mathrm{2}} {x}\centerdot\sqrt{\mathrm{tan}\:{x}}\right){dx}=? \\ $$
Commented by MathematicalUser2357 last updated on 26/Feb/25
$$\left(\mathrm{2}\right)\:\mathrm{Go}\:\mathrm{to}\:\underline{\mathrm{https}://\mathrm{playentry}.\mathrm{org}/\mathrm{project}/\mathrm{67a8a3c0d78abe99420525f0}} \\ $$$$\mathrm{and}\:\mathrm{press}\:\mathrm{the}\:\mathrm{play}\:\mathrm{button}\:\mathrm{and}\:\mathrm{screenshot}\:\mathrm{and}\:\mathrm{give}\:\mathrm{the}\:\mathrm{screenshot}\:\mathrm{to}\:\mathrm{me} \\ $$
Answered by Frix last updated on 26/Feb/25
![∫sec^2 x (√(tan x)) dx =^([t=(√(tan x))]) 2∫t^2 dt=(2/3)t^3 = =(2/3)tan^(3/2) x +C](https://www.tinkutara.com/question/Q217003.png)
$$\int\mathrm{sec}^{\mathrm{2}} \:{x}\:\sqrt{\mathrm{tan}\:{x}}\:{dx}\:\overset{\left[{t}=\sqrt{\mathrm{tan}\:{x}}\right]} {=}\:\mathrm{2}\int{t}^{\mathrm{2}} {dt}=\frac{\mathrm{2}}{\mathrm{3}}{t}^{\mathrm{3}} = \\ $$$$=\frac{\mathrm{2}}{\mathrm{3}}\mathrm{tan}^{\frac{\mathrm{3}}{\mathrm{2}}} \:{x}\:+{C} \\ $$
Commented by MathematicalUser2357 last updated on 27/Feb/25
$$\mathrm{what}\:\mathrm{about}\:\mathrm{the}\:\mathrm{second}\:\mathrm{question}? \\ $$
Answered by MATHEMATICSAM last updated on 27/Feb/25
$$\mathrm{Let}\:\mathrm{tan}{x}\:=\:{t} \\ $$$$\Rightarrow\:\mathrm{sec}^{\mathrm{2}} {x}\:{dx}\:=\:{dt} \\ $$$$ \\ $$$$\int\:\left(\mathrm{sec}^{\mathrm{2}} {x}\sqrt{\mathrm{tan}{x}}\right)\:{dx} \\ $$$$=\:\int\:\sqrt{{t}}\:{dt} \\ $$$$=\:\frac{\mathrm{2}}{\mathrm{3}}\:{t}^{\frac{\mathrm{3}}{\mathrm{2}}} \:+\:{C} \\ $$$$=\:\frac{\mathrm{2}}{\mathrm{3}}\:\mathrm{tan}^{\frac{\mathrm{3}}{\mathrm{2}}} {x}\:+\:{C} \\ $$