Menu Close

dx-a-btanx-

Tinku Tara 0 Comments
Pin




Question Number 189923 by alcohol last updated on 24/Mar/23
∫(dx/(a + btanx))
$$\int\frac{{dx}}{{a}\:+\:{btanx}} \\ $$
Answered by CElcedricjunior last updated on 24/Mar/23
∫(dx/(a+btanx))=k posons tanx=t=>dx=(1/(1+t^2 ))dt k=∫(dt/((a+bt)(1+t^2 ))) k=∫((m/(a+bt))+((nt+c)/(1+t^2 )))dt onr { ((m+nb=0(1))),((m+ac=1(3))) :}na+cb=0(2) (1)−(3)=>nb−ac=−1 𝚫=−a^2 −b^2 𝚫_n =b;𝚫_c =−a =>n=−(b/(a^2 +b^2 ));c=(a/(a^2 +b^2 ))■ Moivre m=(b^2 /(a^2 +b^2 )) k=∫(((b^2 /(a^2 +b^2 ))/(a+bt))+((− (b/(a^2 +b^2 ))t+(a/(a^2 +b^2 )))/(1+t^2 )))dt★ cedric k=(b/(a^2 +b^2 ))ln(a+bt)−(b/(2(a^2 +b^2 )))ln(1+t^2 )+(a/(a^2 +b^2 ))arctan(t)+cste ★junior ∫(dx/(a+btanx))=(b/(a^2 +b^2 ))ln∣acosx+bsinx∣+(a/(a^2 +b^2 ))x+k
$$\int\frac{\boldsymbol{{dx}}}{\boldsymbol{{a}}+\boldsymbol{{btanx}}}=\boldsymbol{{k}} \\ $$$$\boldsymbol{{posons}}\:\boldsymbol{{tanx}}=\boldsymbol{{t}}=>\boldsymbol{{dx}}=\frac{\mathrm{1}}{\mathrm{1}+\boldsymbol{{t}}^{\mathrm{2}} }\boldsymbol{{dt}} \\ $$$$\boldsymbol{{k}}=\int\frac{\boldsymbol{{dt}}}{\left(\boldsymbol{{a}}+\boldsymbol{{bt}}\right)\left(\mathrm{1}+\boldsymbol{{t}}^{\mathrm{2}} \right)} \\ $$$$\boldsymbol{{k}}=\int\left(\frac{\boldsymbol{{m}}}{\boldsymbol{{a}}+\boldsymbol{{bt}}}+\frac{\boldsymbol{{nt}}+\boldsymbol{{c}}}{\mathrm{1}+\boldsymbol{{t}}^{\mathrm{2}} }\right)\boldsymbol{{dt}} \\ $$$$\boldsymbol{{onr}}\:\begin{cases}{\boldsymbol{{m}}+\boldsymbol{{nb}}=\mathrm{0}\left(\mathrm{1}\right)}\\{\boldsymbol{{m}}+\boldsymbol{{ac}}=\mathrm{1}\left(\mathrm{3}\right)}\end{cases}\boldsymbol{{na}}+\boldsymbol{{cb}}=\mathrm{0}\left(\mathrm{2}\right) \\ $$$$\left(\mathrm{1}\right)−\left(\mathrm{3}\right)=>\boldsymbol{{nb}}−\boldsymbol{{ac}}=−\mathrm{1} \\ $$$$\boldsymbol{\Delta}=−\boldsymbol{{a}}^{\mathrm{2}} −\boldsymbol{{b}}^{\mathrm{2}} \\ $$$$\boldsymbol{\Delta}_{\boldsymbol{{n}}} =\boldsymbol{{b}};\boldsymbol{\Delta}_{\boldsymbol{{c}}} =−\boldsymbol{{a}} \\ $$$$=>\boldsymbol{{n}}=−\frac{\boldsymbol{{b}}}{\boldsymbol{{a}}^{\mathrm{2}} +\boldsymbol{{b}}^{\mathrm{2}} };\boldsymbol{{c}}=\frac{\boldsymbol{{a}}}{\boldsymbol{{a}}^{\mathrm{2}} +\boldsymbol{{b}}^{\mathrm{2}} }\blacksquare\:\mathrm{M}\boldsymbol{{oivre}} \\ $$$$\boldsymbol{{m}}=\frac{\boldsymbol{{b}}^{\mathrm{2}} }{\boldsymbol{{a}}^{\mathrm{2}} +\boldsymbol{{b}}^{\mathrm{2}} } \\ $$$$\boldsymbol{{k}}=\int\left(\frac{\frac{\boldsymbol{{b}}^{\mathrm{2}} }{\boldsymbol{{a}}^{\mathrm{2}} +\boldsymbol{{b}}^{\mathrm{2}} }}{\boldsymbol{{a}}+\boldsymbol{{bt}}}+\frac{−\:\frac{\boldsymbol{{b}}}{\boldsymbol{{a}}^{\mathrm{2}} +\boldsymbol{{b}}^{\mathrm{2}} }\boldsymbol{{t}}+\frac{\boldsymbol{{a}}}{\boldsymbol{{a}}^{\mathrm{2}} +\boldsymbol{{b}}^{\mathrm{2}} }}{\mathrm{1}+\boldsymbol{{t}}^{\mathrm{2}} }\right)\boldsymbol{{dt}}\bigstar\:\boldsymbol{{cedric}} \\ $$$$\boldsymbol{{k}}=\frac{\boldsymbol{{b}}}{\boldsymbol{{a}}^{\mathrm{2}} +\boldsymbol{{b}}^{\mathrm{2}} }\boldsymbol{\mathrm{ln}}\left(\boldsymbol{\mathrm{a}}+\boldsymbol{\mathrm{bt}}\right)−\frac{\boldsymbol{\mathrm{b}}}{\mathrm{2}\left(\boldsymbol{\mathrm{a}}^{\mathrm{2}} +\boldsymbol{\mathrm{b}}^{\mathrm{2}} \right)}\boldsymbol{\mathrm{ln}}\left(\mathrm{1}+\boldsymbol{\mathrm{t}}^{\mathrm{2}} \right)+\frac{\boldsymbol{\mathrm{a}}}{\boldsymbol{\mathrm{a}}^{\mathrm{2}} +\boldsymbol{\mathrm{b}}^{\mathrm{2}} }\boldsymbol{\mathrm{arctan}}\left(\boldsymbol{\mathrm{t}}\right)+\boldsymbol{\mathrm{cste}}\:\bigstar\boldsymbol{{junior}} \\ $$$$\int\frac{\boldsymbol{{dx}}}{\boldsymbol{{a}}+\boldsymbol{{btanx}}}=\frac{\boldsymbol{\mathrm{b}}}{\boldsymbol{\mathrm{a}}^{\mathrm{2}} +\boldsymbol{\mathrm{b}}^{\mathrm{2}} }\boldsymbol{\mathrm{ln}}\mid\boldsymbol{{acosx}}+\boldsymbol{{bsinx}}\mid+\frac{\boldsymbol{\mathrm{a}}}{\boldsymbol{\mathrm{a}}^{\mathrm{2}} +\boldsymbol{\mathrm{b}}^{\mathrm{2}} }\boldsymbol{\mathrm{x}}+\boldsymbol{\mathrm{k}} \\ $$$$ \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *