∫ ((5+2cos 2x)/(4+2sin 2x)) dx


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Question Number 137285 by bemath last updated on 31/Mar/21

$\int \frac{5 + 2 \cos 2 x}{4 + 2 \sin 2 x} dx$
	Answered by EDWIN88 last updated on 31/Mar/21

	$E = \int \frac{2 \cos 2 x}{4 + 2 \sin 2 x} dx + \int \frac{5}{4 + 2 \sin 2 x} dx$ $E = \frac{1}{2} \int \frac{d (4 + 2 \sin 2 x)}{4 + 2 \sin 2 x} + \int \frac{5}{4 + 2 \sin 2 x} dx$ $E = \frac{1}{2} \ln ∣ 4 + 2 \sin 2 x ∣ + \int \frac{5}{4 + 2 \sin 2 x} dx$
	Commented by Ar Brandon last updated on 31/Mar/21

	$I = \int \frac{5}{4 + 2 \sin 2 x} dx = \frac{5}{4} \int \frac{dx}{1 + sinxcosx}$ $= \frac{5}{4} \int \frac{\sec^{2} x}{\sec^{2} x + tanx} dx = \frac{5}{4} \int \frac{dt}{t^{2} + t + 1}$ $= \frac{5}{4} \cdot \frac{2}{\sqrt{3}} \arctan (\frac{2 t + 1}{\sqrt{3}}) = \frac{5 \sqrt{3}}{6} \arctan (\frac{2 tanx + 1}{\sqrt{3}}) + C$
	Answered by mathmax by abdo last updated on 31/Mar/21
	$I=∫ ((2cos(2x)+5)/(2sin(2x)+4))dx ⇒I=_(2x=t) (1/2)∫ ((2cost+5)/(2sint +4))dt =_(tan((t/2))=y) (1/2)∫ ((2((1−y^2 )/(1+y^2 ))+5)/(((4y)/(y^2 +1)) +4))((2dy)/(1+y^2 )) =∫ ((2−2y^2 +5+5y^2 )/(4y+4y^2 +4))(dy/(1+y^2 )) =(1/4)∫ ((3y^2 +7)/((y^2 +1)(y^2 +y+1)))dy let decompose F(y)=((3y^2 +7)/((y^2 +1)(y^2 +y+1))) F(y)=((ay+b)/(y^2 +1)) +((my+n)/(y^2 +y+1)) ⇒(ay+b)(y^2 +y+1)+(my+n)(y^2 +1)=3y^2 +7 ⇒ay^3 +ay^2 +ay +by^2 +by +b+my^3 +my +ny^2 +n =3y^2 +7 ⇒ (a+m)y^3 +(a+b+n)y^2 +(a+b)y +b+n =3y^2 +7 ⇒ m=−a ,b=3 ,a=−3 ,n=7−3=4 ⇒ F(y)=((−3y+3)/(y^2 +1)) +((3y+4)/(y^2 +y+1)) ⇒ ∫ F(y)dy =∫((−3y)/(y^2 +1))dy +3∫ (dy/(y^2 +1)) +3∫ ((y+(4/3))/(y^2 +y+1))dy =−(3/2)log(y^2 +1)+3arctany +(3/2)∫ ((2y+1+(8/3)−1)/(y^2 +y+1))dy =−(3/2)log(y^2 +1)+3arctany +(3/2)log(y^2 +y+1)+(5/2)∫ (dy/(y^2 +y+1)) ∫ (dy/(y^2 +y+1))=∫ (dy/((y+(1/2))^2 +(3/4)))=_(y+(1/2)=((√3)/2)z) ((√3)/2) .(4/3) ∫ (dz/(z^2 +1)) =(2/( (√3)))arctan(((2y+1)/( (√3)))) ⇒ I =(1/4){−(3/2)log(y^2 +1)+3arctany +(3/2)log(y^2 +y+1)+(5/( (√3)))arctan(((2y+1)/( (√3))))} +C but y=tanx ⇒ I =−(3/8)log(1+tan^2 x)+(3/4)x +(3/2)log(1+tanx +tan^2 x) +(5/( (√3)))arctan(((2tanx+1)/( (√3)))) +C$
	$I = \int \frac{2 \cos (2 x) + 5}{2 \sin (2 x) + 4} dx \Rightarrow I =_{2 x = t} \frac{1}{2} \int \frac{2 cost + 5}{2 sint + 4} dt$ $=_{\tan (\frac{t}{2}) = y} \frac{1}{2} \int \frac{2 \frac{1 - y^{2}}{1 + y^{2}} + 5}{\frac{4 y}{y^{2} + 1} + 4} \frac{2 dy}{1 + y^{2}} = \int \frac{2 - {2 y}^{2} + 5 + {5 y}^{2}}{4 y + {4 y}^{2} + 4} \frac{dy}{1 + y^{2}}$ $= \frac{1}{4} \int \frac{{3 y}^{2} + 7}{(y^{2} + 1) (y^{2} + y + 1)} dy let decompose F (y) = \frac{{3 y}^{2} + 7}{(y^{2} + 1) (y^{2} + y + 1)}$ $F (y) = \frac{ay + b}{y^{2} + 1} + \frac{my + n}{y^{2} + y + 1} \Rightarrow (ay + b) (y^{2} + y + 1) + (my + n) (y^{2} + 1) = {3 y}^{2} + 7$ $\Rightarrow {ay}^{3} + {ay}^{2} + ay + {by}^{2} + by + b + {my}^{3} + my + {ny}^{2} + n = {3 y}^{2} + 7 \Rightarrow$ $(a + m) y^{3} + (a + b + n) y^{2} + (a + b) y + b + n = {3 y}^{2} + 7 \Rightarrow$ $m = - a, b = 3, a = - 3, n = 7 - 3 = 4 \Rightarrow$ $F (y) = \frac{- 3 y + 3}{y^{2} + 1} + \frac{3 y + 4}{y^{2} + y + 1} \Rightarrow$ $\int F (y) dy = \int \frac{- 3 y}{y^{2} + 1} dy + 3 \int \frac{dy}{y^{2} + 1} + 3 \int \frac{y + \frac{4}{3}}{y^{2} + y + 1} dy$ $= - \frac{3}{2} \log (y^{2} + 1) + 3 arctany + \frac{3}{2} \int \frac{2 y + 1 + \frac{8}{3} - 1}{y^{2} + y + 1} dy$ $= - \frac{3}{2} \log (y^{2} + 1) + 3 arctany + \frac{3}{2} \log (y^{2} + y + 1) + \frac{5}{2} \int \frac{dy}{y^{2} + y + 1}$ $\int \frac{dy}{y^{2} + y + 1} = \int \frac{dy}{{(y + \frac{1}{2})}^{2} + \frac{3}{4}} =_{y + \frac{1}{2} = \frac{\sqrt{3}}{2} z} \frac{\sqrt{3}}{2} . \frac{4}{3} \int \frac{dz}{z^{2} + 1}$ $= \frac{2}{\sqrt{3}} \arctan (\frac{2 y + 1}{\sqrt{3}}) \Rightarrow$ $I = \frac{1}{4} {- \frac{3}{2} \log (y^{2} + 1) + 3 arctany + \frac{3}{2} \log (y^{2} + y + 1) + \frac{5}{\sqrt{3}} \arctan (\frac{2 y + 1}{\sqrt{3}})} + C$ $but y = tanx \Rightarrow$ $I = - \frac{3}{8} \log (1 + \tan^{2} x) + \frac{3}{4} x + \frac{3}{2} \log (1 + tanx + \tan^{2} x)$ $+ \frac{5}{\sqrt{3}} \arctan (\frac{2 tanx + 1}{\sqrt{3}}) + C$
	Commented by mathmax by abdo last updated on 31/Mar/21

	$sorryI = - \frac{3}{8} \log (1 + \tan^{2} x) + \frac{3}{4} x + \frac{3}{8} \log (1 + tanx + \tan^{2} x)$ $+ \frac{5}{4 \sqrt{3}} \arctan (\frac{2 tanx + 1}{\sqrt{3}}) + C$
	Answered by MJS_new last updated on 01/Apr/21

	$\int \frac{5 + 2 \cos 2 x}{4 + 2 \sin 2 x} d x =$ $[t = \tan x \to d x = \frac{d t}{t^{2} + 1}]$ $= \int \frac{3 t^{2} + 7}{4 (t^{2} + 1) (t^{2} + t + 1)} d t =$ $= - \int \frac{t}{t^{2} + 1} d t + \int \frac{2 t + 1}{2 (t^{2} + t + 1)} d t + \int \frac{5 d t}{4 (t^{2} + t + 1)} =$ $= - \frac{1}{2} \ln (t^{2} + 1) + \frac{1}{2} \ln (t^{2} + t + 1) + \frac{5 \sqrt{3}}{6} \arctan \frac{2 t + 1}{\sqrt{3}} =$ $= \frac{1}{2} \ln \frac{t^{2} + t + 1}{t^{2} + 1} + \frac{5 \sqrt{3}}{6} \arctan \frac{2 t + 1}{\sqrt{3}} =$ $= \frac{1}{2} \ln (2 + \sin 2 x) + \frac{5 \sqrt{3}}{6} \arctan \frac{1 + 2 \tan x}{\sqrt{3}} + C$
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