∫ (((3sin x+2))/((2sin x+3)^2 )) dx =?


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Question Number 137249 by bemath last updated on 31/Mar/21

$\int \frac{(3 \sin x + 2)}{{(2 \sin x + 3)}^{2}} dx = ?$
	Answered by bemath last updated on 31/Mar/21
	$let tan ((x/2))=u { ((dx=(2/(u^2 +1)) du)),((sin x=((2u)/(u^2 +1)))) :} I=∫ (((((6u+2u^2 +2)/(u^2 +1))))/((((4u+3u^2 +3)/(u^2 +1)))^2 )) ((2/(u^2 +1)))du I= ∫ ((4(u^2 +3u+1))/((3u^2 +4u+3)^2 )) du ....$
	$let \tan (\frac{x}{2}) = u {\begin{cases} dx = \frac{2}{u^{2} + 1} du \\ \sin x = \frac{2 u}{u^{2} + 1} \end{cases}$ $I = \int \frac{(\frac{6 u + {2 u}^{2} + 2}{u^{2} + 1})}{{(\frac{4 u + {3 u}^{2} + 3}{u^{2} + 1})}^{2}} (\frac{2}{u^{2} + 1}) du$ $I = \int \frac{4 (u^{2} + 3 u + 1)}{{({3 u}^{2} + 4 u + 3)}^{2}} du$ $. . . .$
	Answered by Ar Brandon last updated on 31/Mar/21
	$I=∫((3sinx+2)/((2sinx+3)^2 ))dx=(3/2)∫((2sinx+3−(5/3))/((2sinx+3)^2 ))dx =(3/2)∫(dx/(2sinx+3))−(5/2)∫(dx/((2sinx+3)^2 )) =3∫(dt/((2((2t)/(1+t^2 ))+3)(t^2 +1)))−5∫(dt/((2((2t)/(1+t^2 ))+3)^2 (t^2 +1))) =3∫(dt/(3t^2 +4t+3))−5∫((t^2 +1)/((3t^2 +4t+3)^2 ))dt f(t)=((t^2 +1)/((3t^2 +4t+3)^2 ))=(1/3)(((3t^2 +4t+3−4t)/((3t^2 +4t+3)^2 ))) =(1/(3(3t^2 +4t+3)))−((4t)/(3(3t^2 +4t+3)^2 )) g(t)=((4t)/(3(3t^2 +4t+3)^2 ))=(1/3){(2/3)∙((6t+4)/((3t^2 +4t+3)^2 ))−(8/3)∙(1/((3t^2 +4t+3)^2 ))}$
	$I = \int \frac{3 sinx + 2}{{(2 sinx + 3)}^{2}} dx = \frac{3}{2} \int \frac{2 sinx + 3 - \frac{5}{3}}{{(2 sinx + 3)}^{2}} dx$ $= \frac{3}{2} \int \frac{dx}{2 sinx + 3} - \frac{5}{2} \int \frac{dx}{{(2 sinx + 3)}^{2}}$ $= 3 \int \frac{dt}{(2 \frac{2 t}{1 + t^{2}} + 3) (t^{2} + 1)} - 5 \int \frac{dt}{{(2 \frac{2 t}{1 + t^{2}} + 3)}^{2} (t^{2} + 1)}$ $= 3 \int \frac{dt}{{3 t}^{2} + 4 t + 3} - 5 \int \frac{t^{2} + 1}{{({3 t}^{2} + 4 t + 3)}^{2}} dt$ $f (t) = \frac{t^{2} + 1}{{({3 t}^{2} + 4 t + 3)}^{2}} = \frac{1}{3} (\frac{{3 t}^{2} + 4 t + 3 - 4 t}{{({3 t}^{2} + 4 t + 3)}^{2}})$ $= \frac{1}{3 ({3 t}^{2} + 4 t + 3)} - \frac{4 t}{3 {({3 t}^{2} + 4 t + 3)}^{2}}$ $g (t) = \frac{4 t}{3 {({3 t}^{2} + 4 t + 3)}^{2}} = \frac{1}{3} {\frac{2}{3} \cdot \frac{6 t + 4}{{({3 t}^{2} + 4 t + 3)}^{2}} - \frac{8}{3} \cdot \frac{1}{{({3 t}^{2} + 4 t + 3)}^{2}}}$
	Answered by Dwaipayan Shikari last updated on 31/Mar/21

	$\int \frac{a s i n x + b}{{(b s i n x + a)}^{2}} d x = τ (a, b)$ $= \frac{a^{2}}{b} \int \frac{a b s i n (x) + a^{2}}{{(a b s i n (x) + a^{2})}^{2}} + \frac{b^{2} - a^{2}}{{(a b s i n (x) + a^{2})}^{2}} d x$ $= \frac{a}{b^{2}} \int \frac{1}{s i n (x) + \frac{a}{b}} d x + \frac{(b^{2} - a^{2})}{b^{3}} \int \frac{1}{{(s i n (x) + \frac{a}{b})}^{2}} d x t a n \frac{x}{2} = t$ $J (a) = \int \frac{1}{s i n (x) + \frac{a}{b}} d x = 2 \int \frac{1}{\frac{2 t}{1 + t^{2}} + \frac{a}{b}} . \frac{1}{1 + t^{2}} d t$ $= \frac{2 b}{a} \int \frac{1}{t^{2} + 2 \frac{b}{a} t + 1} d t = \frac{2 b}{a} \int \frac{1}{{(t + \sqrt{\frac{b}{a}})}^{2} + \frac{b - a}{a}} d t$ $= \frac{2 b}{a} \sqrt{\frac{a}{b - a}} {t a n}^{- 1} \frac{\sqrt{a} t + \sqrt{b}}{\sqrt{b - a}} + C$ $J^{'} (a) = - \frac{1}{b} \int \frac{1}{{(s i n x + \frac{a}{b})}^{2}} d x$ $J^{'} (a) = {t a n}^{- 1} \frac{\sqrt{a} t + \sqrt{b}}{\sqrt{b - a}} (- \frac{b {(b - a)}^{- 1 / 2}}{a^{3 / 2}} + \frac{a^{- 1 / 2}}{{(b - a)}^{3 / 2}}) + \frac{b - a}{b - a + {(\sqrt{a} t + \sqrt{b})}^{2}} (\frac{\sqrt{a} t + b}{2 {(b - a)}^{3 / 2}} + \frac{\frac{t}{\sqrt{a}}}{2})$ $. . . τ (a, b) = \frac{a^{2}}{b} J (a) - \frac{b^{2} - a^{2}}{b^{4}} J^{'} (a)$
	Answered by MJS_new last updated on 31/Mar/21

	${(u / v)}^{'} = \frac{u^{'} v - v^{'} u}{v^{2}} \Leftrightarrow \frac{u}{v} = \int \frac{u^{'}}{v} - \frac{v^{'} u}{v^{2}} \Leftrightarrow \int \frac{v^{'} u}{v^{2}} = - \frac{u}{v} + \int \frac{u^{'}}{v}$ $v = 3 + 2 \sin x \Rightarrow v^{'} = 2 \cos x$ $2 u \cos x = 2 + 3 \sin x \Rightarrow u = \frac{2 + 3 \sin x}{2 \cos x} \Rightarrow u^{'} = \frac{3 + 2 \sin x}{{2 \cos}^{2} x}$ $now we have$ $\int \frac{v^{'} u}{v^{2}} = - \frac{u}{v} + \int \frac{u^{'}}{v}$ $\int \frac{2 + 3 \sin x}{{(3 + 2 \sin x)}^{2}} d x = - \frac{2 + 3 \sin x}{2 (3 + 2 \sin x) \cos x} + \frac{1}{2} \int \frac{d x}{\cos^{2} x} =$ $= - \frac{2 + 3 \sin x}{2 (3 + 2 \sin x) \cos x} + \frac{1}{2} \tan x =$ $= - \frac{\cos x}{3 + 2 \sin x} + C$
	Commented by Dwaipayan Shikari last updated on 31/Mar/21

	$S o i t h a s b e e n p r o v e d t h a t e x p e r i e n c e p a y s m o r e : ($
	Commented by MJS_new last updated on 31/Mar/21

	$how old are you ?$ $I^{'} ve been solving integrals for about 35 years . . .$ $this special one I tried using t = \tan \frac{x}{2} first;$ $then I found {Ostrogradski}^{'} s Method leads to$ $\frac{p_{1} (t)}{q_{1} (t)} + \int 0 d t so I knew it can be solved by parts$ $using the quotient rule most people {don}^{'} t$ $know . . .$
	Commented by Dwaipayan Shikari last updated on 31/Mar/21

	$e^{\frac{17}{6}} (E x a c t l y m a y b e) :)$
	Commented by Ar Brandon last updated on 31/Mar/21

	$35 years! {That}^{'} s the sum of our ages$ 😅
	Answered by EDWIN88 last updated on 31/Mar/21

	$E = \int \frac{(2 + 3 \sin x)}{{(3 + 2 \sin x)}^{2}} dx$ $E = \int \frac{2 + 3 \sin x}{2 \cos x} . \frac{2 \cos x}{{(3 + 2 \sin x)}^{2}} dx$ $E = \frac{1}{2} \int \underset{u}{\underset{⏟}{\frac{2 + 3 \sin x}{\cos x}}} . \underset{dv}{\underset{⏟}{\frac{2 \cos x}{{(3 + 2 \sin x)}^{2}} . dx}}$ $by parts$ $\Rightarrow du = \frac{3 + 2 \sin x}{\cos^{2} x} dx; v = - \frac{1}{3 + 2 \sin x}$ $2 E = - \frac{2 + 3 \sin x}{\cos x (3 + 2 \sin x)} + \int \frac{dx}{\cos^{2} x}$ $2 E = - \frac{2 + 3 \sin x}{\cos x (3 + 2 \sin x)} + \tan x + c$ $E = - \frac{2 + 3 \sin x}{2 \cos x (3 + 2 \sin x)} + \frac{\sin x}{2 \cos x} + C$ $E = \frac{- 2 - 3 \sin x + \sin x (3 + 2 \sin x)}{2 \cos x (3 + 2 \sin x)} + C$ $E = \frac{2 \sin^{2} x - 2}{2 \cos x (3 + 2 \sin x)} = \frac{- \cos x}{3 + 2 \sin x} + C$
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