find ∫ (dx/(acosx +bsinx)) with a and b reals


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Question Number 59998 by Mr X pcx last updated on 16/May/19

$f i n d \int \frac{d x}{a c o s x + b s i n x} w i t h a a n d b r e a l s$
Commented by mr W last updated on 16/May/19

$a \cos x + b \sin x = \sqrt{a^{2} + b^{2}} (\frac{a}{\sqrt{a^{2} + b^{2}}} \cos x + \frac{b}{\sqrt{a^{2} + b^{2}}} \sin x)$ $= \sqrt{a^{2} + b^{2}} (\sin θ \cos x + \cos θ \sin x)$ $= \sqrt{a^{2} + b^{2}} \sin (x + θ)$ $w i t h θ = \tan^{- 1} \frac{a}{b}$ $\int \frac{d x}{a c o s x + b s i n x}$ $= \frac{1}{\sqrt{a^{2} + b^{2}}} \int \frac{d x}{s i n (x + θ)}$ $= \frac{1}{2 \sqrt{a^{2} + b^{2}}} \times \ln \frac{1 - \cos (x + θ)}{1 + \cos (x + θ)} + C$
Commented by maxmathsup by imad last updated on 17/May/19

$t h a n k s y o u s i r .$
Commented by maxmathsup by imad last updated on 17/May/19
$let I =∫ (dx/(acosx +bsinx)) changement tan((x/2))=t give I =∫ (1/(a ((1−t^2 )/(1+t^2 )) +b((2t)/(1+t^2 )))) ((2dt)/(1+t^2 )) = ∫ ((2dt)/(a−at^2 +2bt)) =∫ ((−2dt)/(at^2 −2bt−a)) =∫ F(t)dt roots of at^2 −2bt −a =0 →Δ^′ =b^2 +a^2 for a≠0 ⇒Δ^′ >0 ⇒ t_1 =((b+(√(a^2 +b^2 )))/a) and t_2 =((b−(√(a^2 +b^2 )))/a) ⇒F(t) =((−2)/(a(t−t_1 )(t−t_2 ))) =((−2)/(a(t_1 −t_2 ))){ (1/(t−t_1 )) −(1/(t−t_2 ))} =((−2)/(2(√(a^2 +b^2 )))){(1/(t−t_1 )) −(1/(t−t_2 ))} ⇒ I =(1/(√(a^2 +b^2 )))ln∣((t−t_1 )/(t−t_2 ))∣ +C =(1/(√(a^2 +b^2 )))ln∣((tan((x/2))−((b+(√(a^2 +b^2 )))/a))/(tan((x/2))−((b−(√(a^2 +b^2 )))/a)))∣ +C =(1/(√(a^2 +b^2 )))ln∣((atan((x/2))−b−(√(a^2 +b^2 )))/(atan((x/2))−b +(√(a^2 +b^2 ))))∣ +C .$
$l e t I = \int \frac{d x}{a c o s x + b s i n x} c h a n g e m e n t t a n (\frac{x}{2}) = t g i v e$ $I = \int \frac{1}{a \frac{1 - t^{2}}{1 + t^{2}} + b \frac{2 t}{1 + t^{2}}} \frac{2 d t}{1 + t^{2}} = \int \frac{2 d t}{a - {a t}^{2} + 2 b t} = \int \frac{- 2 d t}{{a t}^{2} - 2 b t - a} = \int F (t) d t$ $r o o t s o f {a t}^{2} - 2 b t - a = 0 \to Δ^{^{'}} = b^{2} + a^{2} f o r a \neq 0 \Rightarrow Δ^{^{'}} > 0 \Rightarrow$ $t_{1} = \frac{b + \sqrt{a^{2} + b^{2}}}{a} a n d t_{2} = \frac{b - \sqrt{a^{2} + b^{2}}}{a} \Rightarrow F (t) = \frac{- 2}{a (t - t_{1}) (t - t_{2})}$ $= \frac{- 2}{a (t_{1} - t_{2})} {\frac{1}{t - t_{1}} - \frac{1}{t - t_{2}}} = \frac{- 2}{2 \sqrt{a^{2} + b^{2}}} {\frac{1}{t - t_{1}} - \frac{1}{t - t_{2}}} \Rightarrow$ $I = \frac{1}{\sqrt{a^{2} + b^{2}}} l n ∣ \frac{t - t_{1}}{t - t_{2}} ∣ + C = \frac{1}{\sqrt{a^{2} + b^{2}}} l n ∣ \frac{t a n (\frac{x}{2}) - \frac{b + \sqrt{a^{2} + b^{2}}}{a}}{t a n (\frac{x}{2}) - \frac{b - \sqrt{a^{2} + b^{2}}}{a}} ∣ + C$ $= \frac{1}{\sqrt{a^{2} + b^{2}}} l n ∣ \frac{a t a n (\frac{x}{2}) - b - \sqrt{a^{2} + b^{2}}}{a t a n (\frac{x}{2}) - b + \sqrt{a^{2} + b^{2}}} ∣ + C .$
	Answered by kaivan.ahmadi last updated on 16/May/19

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