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Question-211797

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Question Number 211797 by Spillover last updated on 21/Sep/24
Answered by Ghisom last updated on 21/Sep/24
∫(((√(cot x))−(√(tan x)))/(1+3sin 2x))dx= [t=(√(tan x))] =−2∫((t^2 −1)/(t^4 +6t^2 +1))dt= [u=((2t)/(t^2 +1))] =∫(du/(u^2 +1))=arctan u =arctan ((2t)/(t^2 +1)) = =arctan ((2(√(tan x)))/(1+tan x)) = =arctan ((2(√(cos x sin x)))/(cos x +sin x)) +C
$$\int\frac{\sqrt{\mathrm{cot}\:{x}}−\sqrt{\mathrm{tan}\:{x}}}{\mathrm{1}+\mathrm{3sin}\:\mathrm{2}{x}}{dx}= \\ $$$$\:\:\:\:\:\left[{t}=\sqrt{\mathrm{tan}\:{x}}\right] \\ $$$$=−\mathrm{2}\int\frac{{t}^{\mathrm{2}} −\mathrm{1}}{{t}^{\mathrm{4}} +\mathrm{6}{t}^{\mathrm{2}} +\mathrm{1}}{dt}= \\ $$$$\:\:\:\:\:\left[{u}=\frac{\mathrm{2}{t}}{{t}^{\mathrm{2}} +\mathrm{1}}\right] \\ $$$$=\int\frac{{du}}{{u}^{\mathrm{2}} +\mathrm{1}}=\mathrm{arctan}\:{u}\:=\mathrm{arctan}\:\frac{\mathrm{2}{t}}{{t}^{\mathrm{2}} +\mathrm{1}}\:= \\ $$$$=\mathrm{arctan}\:\frac{\mathrm{2}\sqrt{\mathrm{tan}\:{x}}}{\mathrm{1}+\mathrm{tan}\:{x}}\:= \\ $$$$=\mathrm{arctan}\:\frac{\mathrm{2}\sqrt{\mathrm{cos}\:{x}\:\mathrm{sin}\:{x}}}{\mathrm{cos}\:{x}\:+\mathrm{sin}\:{x}}\:+{C} \\ $$
Answered by TonyCWX08 last updated on 21/Sep/24
I=∫(((√(cot (x)))−(√(tan (x))))/(1+3sin (2x)))dx Using Weierstrass Substitution, let t = tan (x/2) ⇒ x = 2tan^(−1) (t) dx = ((2 dt)/(1+t^2 )) cot(x)=((1−t^2 )/(2t)) tan(x)=((2t)/(1−t^2 )) sin(2x) =2sin (x)cos (x) =2(((2t)/(1+t^2 )))(((1−t^2 )/(1+t^2 ))) =((4t−4t^3 )/((1+t^2 )^2 )) I=∫(((√(((1−t^2 )/(2t)) )) − (√((2t)/(1−t^2 ))))/((4t−4t^3 )/((1+t^2 )^2 )))(((2 dt)/(1+t^2 ))) This can be simplified to I=∫((−t^4 −2t^3 −2t+1)/( (2t−2t^3 )^(3/2) ))dt Can someone continue?
$${I}=\int\frac{\sqrt{\mathrm{cot}\:\left({x}\right)}−\sqrt{\mathrm{tan}\:\left({x}\right)}}{\mathrm{1}+\mathrm{3sin}\:\left(\mathrm{2}{x}\right)}{dx} \\ $$$${Using}\:{Weierstrass}\:{Substitution}, \\ $$$${let}\:{t}\:=\:\mathrm{tan}\:\frac{{x}}{\mathrm{2}}\:\Rightarrow\:{x}\:=\:\mathrm{2tan}^{−\mathrm{1}} \left({t}\right) \\ $$$${dx}\:=\:\frac{\mathrm{2}\:{dt}}{\mathrm{1}+{t}^{\mathrm{2}} } \\ $$$$ \\ $$$${cot}\left({x}\right)=\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{2}{t}} \\ $$$${tan}\left({x}\right)=\frac{\mathrm{2}{t}}{\mathrm{1}−{t}^{\mathrm{2}} } \\ $$$${sin}\left(\mathrm{2}{x}\right) \\ $$$$=\mathrm{2sin}\:\left({x}\right)\mathrm{cos}\:\left({x}\right) \\ $$$$=\mathrm{2}\left(\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}} }\right)\left(\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }\right) \\ $$$$=\frac{\mathrm{4}{t}−\mathrm{4}{t}^{\mathrm{3}} }{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$$$ \\ $$$${I}=\int\frac{\sqrt{\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{2}{t}}\:}\:−\:\sqrt{\frac{\mathrm{2}{t}}{\mathrm{1}−{t}^{\mathrm{2}} }}}{\frac{\mathrm{4}{t}−\mathrm{4}{t}^{\mathrm{3}} }{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{\mathrm{2}} }}\left(\frac{\mathrm{2}\:{dt}}{\mathrm{1}+{t}^{\mathrm{2}} }\right) \\ $$$${This}\:{can}\:{be}\:{simplified}\:{to} \\ $$$${I}=\int\frac{−{t}^{\mathrm{4}} −\mathrm{2}{t}^{\mathrm{3}} −\mathrm{2}{t}+\mathrm{1}}{\:\left(\mathrm{2}{t}−\mathrm{2}{t}^{\mathrm{3}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} }{dt} \\ $$$${Can}\:{someone}\:{continue}? \\ $$

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