∫(dx/(a + btanx))


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Question Number 189923 by alcohol last updated on 24/Mar/23

$\int \frac{d x}{a + b t a n x}$
	Answered by CElcedricjunior last updated on 24/Mar/23
	$∫(dx/(a+btanx))=k posons tanx=t=>dx=(1/(1+t^2 ))dt k=∫(dt/((a+bt)(1+t^2 ))) k=∫((m/(a+bt))+((nt+c)/(1+t^2 )))dt onr { ((m+nb=0(1))),((m+ac=1(3))) :}na+cb=0(2) (1)−(3)=>nb−ac=−1 𝚫=−a^2 −b^2 𝚫_n =b;𝚫_c =−a =>n=−(b/(a^2 +b^2 ));c=(a/(a^2 +b^2 ))■ Moivre m=(b^2 /(a^2 +b^2 )) k=∫(((b^2 /(a^2 +b^2 ))/(a+bt))+((− (b/(a^2 +b^2 ))t+(a/(a^2 +b^2 )))/(1+t^2 )))dt★ cedric k=(b/(a^2 +b^2 ))ln(a+bt)−(b/(2(a^2 +b^2 )))ln(1+t^2 )+(a/(a^2 +b^2 ))arctan(t)+cste ★junior ∫(dx/(a+btanx))=(b/(a^2 +b^2 ))ln∣acosx+bsinx∣+(a/(a^2 +b^2 ))x+k$
	$\int \frac{d x}{a + b t a n x} = k$ $p o s o n s t a n x = t => d x = \frac{1}{1 + t^{2}} d t$ $k = \int \frac{d t}{(a + b t) (1 + t^{2})}$ $k = \int (\frac{m}{a + b t} + \frac{n t + c}{1 + t^{2}}) d t$ $o n r {\begin{cases} m + n b = 0 (1) \\ m + a c = 1 (3) \end{cases} n a + c b = 0 (2)$ $(1) - (3) => n b - a c = - 1$ $Δ = - a^{2} - b^{2}$ $Δ_{n} = b; Δ_{c} = - a$ $=> n = - \frac{b}{a^{2} + b^{2}}; c = \frac{a}{a^{2} + b^{2}} ◼ M o i v r e$ $m = \frac{b^{2}}{a^{2} + b^{2}}$ $k = \int (\frac{\frac{b^{2}}{a^{2} + b^{2}}}{a + b t} + \frac{- \frac{b}{a^{2} + b^{2}} t + \frac{a}{a^{2} + b^{2}}}{1 + t^{2}}) d t ★ c e d r i c$ $k = \frac{b}{a^{2} + b^{2}} \ln (a + bt) - \frac{b}{2 (a^{2} + b^{2})} \ln (1 + t^{2}) + \frac{a}{a^{2} + b^{2}} \arctan (t) + cste ★ j u n i o r$ $\int \frac{d x}{a + b t a n x} = \frac{b}{a^{2} + b^{2}} \ln ∣ a c o s x + b s i n x ∣ + \frac{a}{a^{2} + b^{2}} x + k$
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