(1) ∫ (dx/(x^6 −64)) (2) y′ + y tan x = sin 2x y(0) = 1


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Question Number 105039 by bemath last updated on 25/Jul/20

$(1) \int \frac{d x}{x^{6} - 64}$ $(2) y^{'} + y \tan x = \sin 2 x$ $y (0) = 1$
	Answered by bobhans last updated on 25/Jul/20
	$(2) If u(x) = e^(∫ tan (x) dx) = e^(ln ((1/(cos x))) ) =(1/(cos x))=sec x y(x) = ((∫ sin 2x.sec x dx + C)/(sec x)) y(x) = cos x.{∫2sin x dx + C } y(x) = cos x. {−2cos x + C } y(x) = −2cos^2 x + C.cos x ★ y(0) = −2+C = 1 ⇔C = 3 y(x) = −2cos^2 x+3cos x$
	$(2) I f u (x) = e^{\int \tan (x) d x} = e^{\ln (\frac{1}{\cos x})} = \frac{1}{\cos x} = \sec x$ $y (x) = \frac{\int \sin 2 x . \sec x d x + C}{\sec x}$ $y (x) = \cos x . {\int 2 \sin x d x + C}$ $y (x) = \cos x . {- 2 \cos x + C}$ $y (x) = - 2 \cos^{2} x + C . \cos x ★$ $y (0) = - 2 + C = 1 \Leftrightarrow C = 3$ $y (x) = - 2 \cos^{2} x + 3 \cos x$
	Answered by mathmax by abdo last updated on 25/Jul/20

	$1) complex method we decompose F (z) = \frac{1}{z^{6} - 64} \Rightarrow$ $z^{6} = 64 let z = {re}^{i θ} so e \Rightarrow r^{6} e^{i 6 θ} = 64 e^{i (2 k π)} \Rightarrow r =^{6} \sqrt{64} = 2$ $θ = \frac{k π}{3} and k \in [[0, 5]] so the roots are z_{k} = {2 e}^{i \frac{k π}{3}} and 0 ⩽ k ⩽ 5$ $\Rightarrow F (z) = \frac{1}{\prod_{k = 0}^{5} (z - z_{k})} = \sum_{k = 0}^{5} \frac{a_{k}}{z - z_{k}} with a_{k} = \frac{1}{{6 z}_{k}^{5}} = \frac{z_{k}}{6 (64)} = \frac{z_{k}}{384}$ $\Rightarrow F (z) = \frac{1}{384} \sum_{k = 0}^{5} \frac{z_{k}}{z - z_{k}} \Rightarrow \int F (z) dz = \frac{1}{384} \sum_{k = 0}^{5} z_{k} \int \frac{dz}{z - {2 e}^{\frac{ik π}{3}}}$ $= \frac{1}{384} \sum_{k = 0}^{5} ({2 e}^{\frac{ik π}{3}}) \ln (z - {2 e}^{\frac{ik π}{3}}) + C$
	Commented by Dwaipayan Shikari last updated on 25/Jul/20

	$I t h i n k i t c a n b e d o n e w i t h \frac{1}{16} \int \frac{1}{x^{3} - 8} - \int \frac{1}{x^{3} + 8}$ $= \frac{1}{16} \int \frac{A}{x - 2} + \frac{B x + C}{x^{2} + 2 x + 4} - \frac{1}{16} \int \frac{A^{'}}{x + 2} + \frac{B^{'} x + C^{'}}{x^{2} - 2 x + 4}$
	Commented by mathmax by abdo last updated on 26/Jul/20

	$yes its another way for tbis integral$
	Answered by mathmax by abdo last updated on 25/Jul/20
	$y^′ +ytanx =sin(2x) h)→y^′ =−ytanx ⇒(y^′ /y) =−tanx ⇒ln∣y∣ =−∫ ((sinx)/(cosx))dx =ln∣cosx∣ +c ⇒ y(x) =k ∣cosx∣ let find solution on {x /cosx >0} mvc method →y^′ =k^′ cosx −ksinx e⇒k^′ cosx−ksinx +kcosx ×((sinx)/(cosx)) =sin(2x) ⇒k^′ =((sin(2x))/(cosx)) =2sinx ⇒ k =2∫sinx dx =−2cosx +λ ⇒y(x) =(−2cosx +λ)cosx ⇒y(x) =λcosx−sin(2x) y(0)=1 ⇒λ−0 =1 ⇒λ =1 ⇒y(x) =cosx −sin(2x)$
	$y^{^{'}} + ytanx = \sin (2 x)$ $h) \to y^{^{'}} = - ytanx \Rightarrow \frac{y^{^{'}}}{y} = - tanx \Rightarrow \ln ∣ y ∣ = - \int \frac{sinx}{cosx} dx = \ln ∣ cosx ∣ + c \Rightarrow$ $y (x) = k ∣ cosx ∣ let find solution on {x / cosx > 0}$ $mvc method \to y^{^{'}} = k^{^{'}} cosx - ksinx$ $e \Rightarrow k^{^{'}} cosx - ksinx + kcosx \times \frac{sinx}{cosx} = \sin (2 x) \Rightarrow k^{^{'}} = \frac{\sin (2 x)}{cosx} = 2 sinx \Rightarrow$ $k = 2 \int sinx dx = - 2 cosx + λ \Rightarrow y (x) = (- 2 cosx + λ) cosx$ $\Rightarrow y (x) = λ cosx - \sin (2 x)$ $y (0) = 1 \Rightarrow λ - 0 = 1 \Rightarrow λ = 1 \Rightarrow y (x) = cosx - \sin (2 x)$
	Answered by Dwaipayan Shikari last updated on 25/Jul/20

	$\frac{d y}{d x} + y t a n x = s i n 2 x$ $I . F = e^{\int t a n x d x} = e^{l o g (s e c x)} = s e c x$ $y . s e c x = \int s i n 2 x s e c x d x = 2 \int s i n x = - 2 c o s x + C$ $y = - 2 {c o s}^{2} x + C c o s x$ $y (x) = - 2 {c o s}^{2} x + C c o s x$ $y (0) = - 2 + C$ $C = 3$ $y (x) = - 2 {c o s}^{2} x + 3 c o s x ★$
	Answered by OlafThorendsen last updated on 25/Jul/20

	$(1) R (x) = \frac{1}{x^{6} - 1}$ $R (x) = \frac{1}{(x - 1) (x + 1) (x^{2} - x + 1) (x^{2} - x + 1)}$ $\int R (x) d x =$ $\int (\frac{1}{6 (x - 1)} - \frac{1}{6 (x + 1)} + \frac{x - 2}{6 (x^{2} - x + 1)} - \frac{x + 2}{6 (x^{2} + x + 1)}) d x$ $= \frac{1}{6} \ln ∣ \frac{x - 1}{x + 1} ∣$ $+ \frac{1}{12} \int \frac{2 x - 1}{x^{2} - x + 1} d x - \frac{1}{12} \int \frac{2 x + 1}{x^{2} + x + 1} d x$ $- \frac{1}{4} \int \frac{d x}{x^{2} - x + 1} - \frac{1}{4} \int \frac{d x}{x^{2} + x + 1}$ $= \frac{1}{6} \ln ∣ \frac{x - 1}{x + 1} ∣ + \frac{1}{12} \ln ∣ \frac{x^{2} - x + 1}{x^{2} + x + 1} ∣$ $- \frac{1}{3} \int \frac{d x}{\frac{4}{3} {(x - \frac{1}{2})}^{2} + 1} - \frac{1}{3} \int \frac{d x}{\frac{4}{3} {(x + \frac{1}{2})}^{2} + 1}$ $= \frac{1}{6} \ln ∣ \frac{x - 1}{x + 1} ∣ + \frac{1}{12} \ln ∣ \frac{x^{2} - x + 1}{x^{2} + x + 1} ∣$ $- \frac{1}{2 \sqrt{3}} \arctan [\frac{2}{\sqrt{3}} (x - \frac{1}{2})] - \frac{1}{2 \sqrt{3}} \arctan [\frac{2}{\sqrt{3}} (x + \frac{1}{2})]$ $\arctan u + \arctan v = \arctan \frac{u + v}{1 - u v}$ $u = \frac{2}{\sqrt{3}} (x - \frac{1}{2}) and v = \frac{2}{\sqrt{3}} (x + \frac{1}{2})$ $\frac{u + v}{1 - u v} = \frac{\frac{4}{\sqrt{3}} x}{1 - \frac{4}{3} (x^{2} - \frac{1}{4})} = \frac{\sqrt{3} x}{1 - x^{2}}$ $Finally :$ $\int R (x) d x = \frac{1}{6} \ln ∣ \frac{x - 1}{x + 1} ∣ + \frac{1}{12} \ln ∣ \frac{x^{2} - x + 1}{x^{2} + x + 1} ∣ - \frac{1}{2 \sqrt{3}} \arctan \frac{\sqrt{3} x}{1 - x^{2}} + C$
	Commented by bemath last updated on 27/Jul/20

	$t h e q u e s t i o n \int \frac{d x}{x^{6} - 64} s i r$
	Answered by OlafThorendsen last updated on 25/Jul/20

	$Q 2 .$ $y^{'} \cos x - (- y \sin x) = \cos x \sin 2 x = {2 \cos}^{2} x \sin x$ $\frac{y^{'} \cos x - (- y \sin x)}{\cos^{2} x} = 2 \sin x$ $\frac{d}{d x} (\frac{y}{\cos x}) = 2 \sin x$ $\frac{y}{\cos x} = - 2 \cos x + C$ $y = - {2 \cos}^{2} x + Ccos x$ $y (0) = - 2 + C = 1 \Rightarrow C = 3$ $y = 3 \cos x - {2 \cos}^{2} x$
	Answered by bramlex last updated on 26/Jul/20

	$(1) \int \frac{d x}{x^{6} - {(2)}^{6}} = \frac{1}{24} \ln (\frac{{(x - 2)}^{2} (x^{2} - 2 x + 4)}{{(x + 2)}^{2} (x^{2} + 2 x + 4)}) - \frac{1}{4 \sqrt{3}} \tan^{- 1} (\frac{2}{x^{2} + 2}) + C$
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