1-sec-2-x-tan-x-dx-

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

(2) Go to https://playentry.org/project/67a8a3c0d78abe99420525f0 and press the play button and screenshot and give the screenshot to me

$$\left(\mathrm{2}\right)\:\mathrm{Go}\:\mathrm{to}\:\underline{\mathrm{https}://\mathrm{playentry}.\mathrm{org}/\mathrm{project}/\mathrm{67a8a3c0d78abe99420525f0}} \\ $$$$\mathrm{and}\:\mathrm{press}\:\mathrm{the}\:\mathrm{play}\:\mathrm{button}\:\mathrm{and}\:\mathrm{screenshot}\:\mathrm{and}\:\mathrm{give}\:\mathrm{the}\:\mathrm{screenshot}\:\mathrm{to}\:\mathrm{me} \\ $$

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

$$\int\mathrm{sec}^{\mathrm{2}} \:{x}\:\sqrt{\mathrm{tan}\:{x}}\:{dx}\:\overset{\left[{t}=\sqrt{\mathrm{tan}\:{x}}\right]} {=}\:\mathrm{2}\int{t}^{\mathrm{2}} {dt}=\frac{\mathrm{2}}{\mathrm{3}}{t}^{\mathrm{3}} = \\ $$$$=\frac{\mathrm{2}}{\mathrm{3}}\mathrm{tan}^{\frac{\mathrm{3}}{\mathrm{2}}} \:{x}\:+{C} \\ $$

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

Let tanx = t ⇒ sec^2 x dx = dt ∫ (sec^2 x(√(tanx))) dx = ∫ (√t) dt = (2/3) t^(3/2) + C = (2/3) tan^(3/2) x + C

$$\mathrm{Let}\:\mathrm{tan}{x}\:=\:{t} \\ $$$$\Rightarrow\:\mathrm{sec}^{\mathrm{2}} {x}\:{dx}\:=\:{dt} \\ $$$$ \\ $$$$\int\:\left(\mathrm{sec}^{\mathrm{2}} {x}\sqrt{\mathrm{tan}{x}}\right)\:{dx} \\ $$$$=\:\int\:\sqrt{{t}}\:{dt} \\ $$$$=\:\frac{\mathrm{2}}{\mathrm{3}}\:{t}^{\frac{\mathrm{3}}{\mathrm{2}}} \:+\:{C} \\ $$$$=\:\frac{\mathrm{2}}{\mathrm{3}}\:\mathrm{tan}^{\frac{\mathrm{3}}{\mathrm{2}}} {x}\:+\:{C} \\ $$

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