∫(dx/((a+bcosx)^2 ))


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Question Number 116999 by TANMAY PANACEA last updated on 08/Oct/20

$\int \frac{d x}{{(a + b c o s x)}^{2}}$
	Answered by Olaf last updated on 08/Oct/20

	$I (x) = \int \frac{d x}{{(a + b \frac{e^{i x} + e^{- i x}}{2})}^{2}}$ $u = e^{i x}$ $d u = {i e}^{i x} d x = i u d x$ $d x = - i \frac{d u}{u}$ $I (u) = - i \int \frac{\frac{d u}{u}}{{(a + b \frac{u + \frac{1}{u}}{2})}^{2}}$ $I (u) = - i \int \frac{u d u}{{(a u + b \frac{u^{2} + 1}{2})}^{2}}$ $I (u) = - i \int \frac{4 u d u}{{({b u}^{2} + 2 a u + b)}^{2}}$ $I (u) = - \frac{4 i}{b^{2}} \int \frac{4 u d u}{{[{(u + \frac{a}{b})}^{2} + 1 - \frac{a^{2}}{b^{2}}]}^{2}}$ $Different cases in function of sign of$ $1 - \frac{a^{2}}{b^{2}} . . .$
	Answered by mathmax by abdo last updated on 08/Oct/20

	$let φ (a) = \int \frac{dx}{a + bcosx} \Rightarrow φ^{^{'}} (a) = - \int \frac{dx}{{(a + bcosx)}^{2}} \Rightarrow$ $\int \frac{dx}{{(a + bcosx)}^{2}} = - φ^{^{'}} (a) but φ (a) =_{\tan (\frac{x}{2}) = t} \int \frac{2 dt}{(1 + t^{2}) (a + b \frac{1 - t^{2}}{1 + t^{2}})}$ $= \int \frac{2 dt}{a + {at}^{2} + b - {bt}^{2}} = 2 \int \frac{dt}{(a - b) t^{2} + a + b}$ $= \frac{2}{a - b} \int \frac{dt}{t^{2} + \frac{a + b}{a - b}} case 1 \frac{a + b}{a - b} > 0 we do ch . t = \sqrt{\frac{a + b}{a - b}} z \Rightarrow$ $φ (a) = \frac{2}{a - b} \times \frac{a - b}{a + b} \int \frac{1}{1 + z^{2}} \frac{\sqrt{a + b}}{\sqrt{a - b}} dz = \frac{2}{\sqrt{a^{2} - b^{2}}} \arctan (\sqrt{\frac{a - b}{a + b}} t)$ $= \frac{2}{\sqrt{a^{2} - b^{2}}} \arctan (\sqrt{\frac{a - b}{a + b}} \tan (\frac{x}{2})) + c \Rightarrow$ $φ^{^{'}} (a) = 2 {(a^{2} - b^{2})}^{- \frac{1}{2}} = - 2 a {(a^{2} - b^{2})}^{- \frac{3}{2}} \arctan (\sqrt{\frac{a - b}{a + b}} \tan (\frac{x}{2}))$ $+ \frac{2}{\sqrt{a^{2} - b^{2}}} \tan (\frac{x}{2}) \times \frac{{(\frac{a - b}{a + b})}^{^{'}}}{2 \sqrt{\frac{a - b}{a + b}} (1 + \frac{a - b}{a + b} \tan^{2} (\frac{x}{2}))} with$ ${(\frac{a - b}{a + b})}^{^{'}} = \frac{a + b - (a - b)}{{(a + b)}^{2}} = \frac{2 b}{{(a + b)}^{2}}$ $case 2 \frac{a + b}{a - b} < 0 \Rightarrow φ (a) = \frac{2}{a - b} \int \frac{dt}{t^{2} - (\frac{a + b}{b - a})}$ $=_{t = \sqrt{\frac{b + a}{b - a}} u} \frac{2}{a - b} \times \frac{b - a}{b + a} \int \frac{1}{u^{2} - 1} \times \frac{\sqrt{b + a}}{\sqrt{b - a}} du$ $= \frac{- 2}{\sqrt{b^{2} - a^{2}}} \int (\frac{1}{u - 1} - \frac{1}{u + 1}) du = \frac{- 2}{\sqrt{b^{2} - a^{2}}} \ln ∣ \frac{u - 1}{u + 1} ∣ + c$ $= \frac{- 2}{\sqrt{b^{2} - a^{2}}} \ln ∣ \frac{\sqrt{\frac{b - a}{b + a}} \tan (\frac{x}{2}) - 1}{\sqrt{\frac{b - a}{b + a}} \tan (\frac{x}{2}) + 1} ∣ + c$ $rest calculus of φ^{^{'}} (a) . . .$
	Answered by TANMAY PANACEA last updated on 09/Oct/20
	$p=((sinx)/(a+bcosx))→(dp/dx)=(((a+bcosx)cosx−sinx(−bsinx))/((a+bcosx)^2 )) (dp/dx)=((acosx+b)/((a+bcosx)^2 ))=(((a/b)(bcosx+a−a)+b)/((a+bcosx)^2 )) (dp/dx)=((a/b)/((a+bcosx)))+(((b^2 −a^2 )/b)/((a+bcosx)^2 )) ∫d(((sinx)/(a+bcosx)))=(a/b)∫(dx/(b((a/b)+((1−tan^2 (x/2))/(1+tan^2 (x/2))))))+((b^2 −a^2 )/b)×I ((sinx)/(a+bcosx))=(a/b^2 )∫((sec^2 (x/2))/((a/b)+1+((a/b)−1)tan^2 (x/2)))dx+((b^2 −a^2 )/b)×I ((sinx)/(a+bcosx))=(a/b^2 )∫((sec^2 (x/2))/(((a/b)−1)(((a+b)/b)×(b/(a−b))+tan^2 (x/2))))+((b^2 −a^2 )/b)I ((sinx)/(a+bcosx))=(a/b^2 )×(b/((a−b)))∫((d(tan(x/2))×2)/({((√((a+b)/(a−b))) )^2 +tan^2 (x/2)))+((b^2 −a^2 )/b)×I ((b/(b^2 −a^2 )))(((sinx)/(a+bcosx)))=(a/b)×(1/(a−b))×((2b)/((a+b)(a−b)))×(1/( (√((a+b)/(a−b)))))×tan^(−1) (((tan(x/2))/( (√((a+b)/(a−b))))))+I so I= (b/(b^2 −a^2 ))(((sinx)/(a+bcosx)))−((2a)/((a+b)(a−b)^2 ))×((√(a−b))/( (√(a+b))))×tan^(−1) (((tan(x/2))/( (√((a+b)/(a−b))))))+C I=(b/(b^2 −a^2 ))(((sinx)/(a+bcosx)))−((2a)/((a^2 −b^2 )^(3/2) ))tan^(−1) (((tan(x/2))/( (√((a+b)/(a−b))))))+C pls check$
	$p = \frac{s i n x}{a + b c o s x} \to \frac{d p}{d x} = \frac{(a + b c o s x) c o s x - s i n x (- b s i n x)}{{(a + b c o s x)}^{2}}$ $\frac{d p}{d x} = \frac{a c o s x + b}{{(a + b c o s x)}^{2}} = \frac{\frac{a}{b} (b c o s x + a - a) + b}{{(a + b c o s x)}^{2}}$ $\frac{d p}{d x} = \frac{\frac{a}{b}}{(a + b c o s x)} + \frac{\frac{b^{2} - a^{2}}{b}}{{(a + b c o s x)}^{2}}$ $\int d (\frac{s i n x}{a + b c o s x}) = \frac{a}{b} \int \frac{d x}{b (\frac{a}{b} + \frac{1 - {t a n}^{2} \frac{x}{2}}{1 + {t a n}^{2} \frac{x}{2}})} + \frac{b^{2} - a^{2}}{b} \times I$ $\frac{s i n x}{a + b c o s x} = \frac{a}{b^{2}} \int \frac{{s e c}^{2} \frac{x}{2}}{\frac{a}{b} + 1 + (\frac{a}{b} - 1) {t a n}^{2} \frac{x}{2}} d x + \frac{b^{2} - a^{2}}{b} \times I$ $\frac{s i n x}{a + b c o s x} = \frac{a}{b^{2}} \int \frac{{s e c}^{2} \frac{x}{2}}{(\frac{a}{b} - 1) (\frac{a + b}{b} \times \frac{b}{a - b} + {t a n}^{2} \frac{x}{2})} + \frac{b^{2} - a^{2}}{b} I$ $\frac{s i n x}{a + b c o s x} = \frac{a}{b^{2}} \times \frac{b}{(a - b)} \int \frac{d (t a n \frac{x}{2}) \times 2}{{{(\sqrt{\frac{a + b}{a - b}})}^{2} + {t a n}^{2} \frac{x}{2}} + \frac{b^{2} - a^{2}}{b} \times I$ $(\frac{b}{b^{2} - a^{2}}) (\frac{s i n x}{a + b c o s x}) = \frac{a}{b} \times \frac{1}{a - b} \times \frac{2 b}{(a + b) (a - b)} \times \frac{1}{\sqrt{\frac{a + b}{a - b}}} \times {t a n}^{- 1} (\frac{t a n \frac{x}{2}}{\sqrt{\frac{a + b}{a - b}}}) + I$ $s o I =$ $\frac{b}{b^{2} - a^{2}} (\frac{s i n x}{a + b c o s x}) - \frac{2 a}{(a + b) {(a - b)}^{2}} \times \frac{\sqrt{a - b}}{\sqrt{a + b}} \times {t a n}^{- 1} (\frac{t a n \frac{x}{2}}{\sqrt{\frac{a + b}{a - b}}}) + C$ $I = \frac{b}{b^{2} - a^{2}} (\frac{s i n x}{a + b c o s x}) - \frac{2 a}{{(a^{2} - b^{2})}^{\frac{3}{2}}} {t a n}^{- 1} (\frac{t a n \frac{x}{2}}{\sqrt{\frac{a + b}{a - b}}}) + C$ $p l s c h e c k$
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