∫ (dx/(a sin x + b cos x))


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Question Number 120758 by bramlexs22 last updated on 02/Nov/20

$\int \frac{dx}{a \sin x + b \cos x}$
	Answered by Dwaipayan Shikari last updated on 02/Nov/20

	$2 \int \frac{d t}{a (\frac{2 t}{1 + t^{2}}) + b (\frac{1 - t^{2}}{1 + t^{2}})} . \frac{1}{1 + t^{2}} t a n \frac{x}{2} = t$ $= 2 \int \frac{d t}{2 a t + b - {b t}^{2}}$ $= - \frac{2}{b} \int \frac{d t}{t^{2} - \frac{2 a t}{b} - 1}$ $= - \frac{2}{b} \int \frac{d t}{{(t - \frac{a}{b})}^{2} - \frac{a^{2}}{b^{2}} - 1}$ $= \frac{2}{b} \int \frac{d t}{{(\sqrt{\frac{a^{2}}{b^{2}} + 1})}^{2} - {(t - \frac{a}{b})}^{2}}$ $= \frac{2}{b} . \frac{1}{2 \sqrt{\frac{a^{2}}{b^{2}} + 1}} l o g (\frac{t - \frac{a}{b} + \sqrt{\frac{a^{2}}{b^{2}} + 1}}{t - \frac{a}{b} - \sqrt{\frac{a^{2}}{b^{2}} + 1}}) + C$ $= \frac{1}{b \sqrt{\frac{a^{2}}{b^{2}} + 1}} . l o g (\frac{t a n \frac{x}{2} - \frac{a}{b} + \sqrt{\frac{a^{2}}{b^{2}} + 1}}{t a n \frac{x}{2} - \frac{a}{b} - \sqrt{\frac{a^{2}}{b^{2}} + 1}}) + C$ $= \frac{1}{\sqrt{a^{2} + b^{2}}} l o g (\frac{b t a n \frac{x}{2} - a + \sqrt{a^{2} + b^{2}}}{b t a n \frac{x}{2} - a - \sqrt{a^{2} + b^{2}}}) + C$
	Answered by liberty last updated on 02/Nov/20
	$find the two numbers r and β such that a = r cos β , b = r sin β → { ((r =(√(a^2 +b^2 )))),((β=tan^(−1) ((b/a)))) :} ∴ ∫ (dx/(a sin x + b cos x)) = (1/r)∫ (dx/(sin (x+β))) =(1/r) ∫ cosec (x+β) dx = (1/r) ln (tan (((x+β)/2)))+c = (1/( (√(a^2 +b^2 )))) ln (tan (((x+β)/2))) + c =(1/( (√(a^2 +b^2 )))) ln (tan (((tan^(−1) ((b/a))+x)/2))) +c$
	$find the two numbers r and β such that$ $a = r \cos β, b = r \sin β \to {\begin{cases} r = \sqrt{a^{2} + b^{2}} \\ β = \tan^{- 1} (\frac{b}{a}) \end{cases}$ $∴ \int \frac{dx}{a \sin x + b \cos x} = \frac{1}{r} \int \frac{dx}{\sin (x + β)}$ $= \frac{1}{r} \int cosec (x + β) dx = \frac{1}{r} \ln (\tan (\frac{x + β}{2})) + c$ $= \frac{1}{\sqrt{a^{2} + b^{2}}} \ln (\tan (\frac{x + β}{2})) + c$ $= \frac{1}{\sqrt{a^{2} + b^{2}}} \ln (\tan (\frac{\tan^{- 1} (\frac{b}{a}) + x}{2})) + c$
	Answered by 675480065 last updated on 02/Nov/20

	$a s i n x + b c o s x = a (\frac{2 t}{1 + t^{2}}) + b (\frac{1 - t^{2}}{1 + t^{2}})$ $t = t a n (\frac{x}{2}) \Rightarrow 2 d t = {s e c}^{2} (\frac{x}{2}) d x$ $\Rightarrow d x = \frac{2 d t}{1 + t^{2}}$ $I = \int \frac{2 d t}{(1 + t^{2})} . \frac{(1 + t^{2})}{2 a t + b (1 - t^{2})} = 2 \int \frac{d t}{(2 a t - {b t}^{2} + b)}$ $I = - 2 \int \frac{d t}{({b t}^{2} - 2 a t - b)} = - 2 \int \frac{d t}{b (t^{2} - 2 a t - b + {(a)}^{2} - (a^{2}))}$ $\Rightarrow I = - 2 \int \frac{d t}{b [{(t - a)}^{2} - (a^{2} + b)]}$ $\Rightarrow I = - \frac{2}{b} \int \frac{d t}{{(t - a)}^{2} - \sqrt{{(a^{2} + b)}^{2}}}$ $L e t (t - a) = \sqrt{a^{2} + b} t a n h β$ $\Rightarrow d t = \sqrt{a^{2} + b} {s e c h}^{2} β d β$ $\Rightarrow I = - \frac{2}{b} \int \frac{\sqrt{a^{2} + b} {s e c h}^{2} β d β}{\sqrt{{(a^{2} + b)}^{2}} ({t a n h}^{2} β - 1)}$ $\Rightarrow I = - \frac{2}{b \sqrt{a^{2} + b}} \int d β = - 2 (\frac{1}{b \sqrt{a^{2} + b}}) {t a n h}^{- 1} (\frac{t - a}{\sqrt{a^{2} + b}}) + k$ $\Rightarrow I = - \frac{1}{b \sqrt{a^{2} + b}} l n ∣ \frac{t - a + \sqrt{a^{2} + b}}{t - a - \sqrt{a^{2} + b}} ∣ + k$
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