∫_0 ^( π/2) ((cos^2 x)/((2cos x+sin x)^2 )) dx =?


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Question Number 144530 by EDWIN88 last updated on 26/Jun/21

$\int_{0}^{π / 2} \frac{\cos^{2} x}{{(2 \cos x + \sin x)}^{2}} dx = ?$
	Answered by mathmax by abdo last updated on 26/Jun/21

	$Φ = \int_{0}^{\frac{π}{2}} \frac{\cos^{2} x}{{(2 cosx + sinx)}^{2}} dx \Rightarrow Φ = \int_{0}^{\frac{π}{2}} \frac{\cos^{2} x}{{4 \cos}^{2} (x) + 4 cosxsinx + \sin^{2} x} dx$ $= \int_{0}^{\frac{π}{2}} \frac{\cos^{2} x}{{3 \cos}^{2} x + 4 cosxsinx + 1} dx = \int_{0}^{\frac{π}{2}} \frac{1 + \cos (2 x)}{2 (3 \frac{1 + \cos (2 x)}{2} + 2 \sin (2 x) + 1)} dx$ $= \int_{0}^{\frac{π}{2}} \frac{1 + \cos (2 x)}{3 + 3 \cos (2 x) + 4 \sin (2 x) + 2} dx =_{2 x = t}$ $= \frac{1}{2} \int_{0}^{π} \frac{1 + cost}{3 + 3 cost + 4 sint + 2} dt$ $=_{\tan (\frac{t}{2}) = y} \frac{1}{2} \int_{0}^{\infty} \frac{1 + \frac{1 - y^{2}}{1 + y^{2}}}{3 + 3 \frac{1 - y^{2}}{1 + y^{2}} + 4 . \frac{2 y}{1 + y^{2}} + 2} \times \frac{2 dy}{1 + y^{2}}$ $= \int_{0}^{\infty} \frac{2}{(3 + {3 y}^{2} + 3 - {3 y}^{2} + 8 y + 2 + {2 y}^{2}) (1 + y^{2})} dy$ $= \int_{0}^{\infty} \frac{2 dy}{(1 + y^{2}) (8 + {2 y}^{2} + 8 y)} dy = \int_{0}^{\infty} \frac{dy}{(y^{2} + 1) (y^{2} + 4 y + 4)}$ $= \int_{0}^{\infty} \frac{dy}{{(y + 2)}^{2} (y^{2} + 1)}$ $F (y) = \frac{1}{{(y + 2)}^{2} (y^{2} + 1)} = \frac{a}{y + 2} + \frac{b}{{(y + 2)}^{2}} + \frac{cy + d}{y^{2} + 1}$ $b = \frac{1}{5} the determination of coefficient is not bad \Rightarrow$ $\int F (y) dy = alog ∣ y + 2 ∣ - \frac{b}{y + 2} + \frac{c}{2} \log (y^{2} + 1) + darctany + C$ $Φ = alog ∣ \tan (x) + 2 ∣ - \frac{b}{tanx + 2} + \frac{c}{2} \log (1 + \tan^{2} x) + d x + C$
	Answered by Ar Brandon last updated on 26/Jun/21

	$I = \int_{0}^{\frac{π}{2}} \frac{\cos^{2} x}{{(2 cosx + sinx)}^{2}} dx = \int_{0}^{\frac{π}{2}} \frac{dx}{{(2 + tanx)}^{2}}$ $= \int_{0}^{\infty} \frac{2 dt}{{(2 + t)}^{2} (1 + t^{2})} = 2 \int_{0}^{\infty} \frac{dt}{{(t + 2)}^{2} (t^{2} + 1)}$ $= 2 \int_{0}^{\infty} (\frac{4}{25} \cdot \frac{1}{t + 2} + \frac{1}{5} \cdot \frac{1}{{(t + 2)}^{2}} - \frac{1}{25} \cdot \frac{4 t - 3}{t^{2} + 1}) dt$ $= 2 {[\frac{4}{25} \ln (t + 2) - \frac{1}{5} \cdot \frac{1}{t + 2} - \frac{2}{25} \ln (t^{2} + 1) + \frac{3}{25} \arctan (t)]}_{0}^{\infty}$ $= {[\frac{4}{25} \ln (\frac{t^{2} + 2 t + 2}{t^{2} + 1}) - \frac{2}{5 (t + 2)} + \frac{6}{25} \arctan (t)]}_{0}^{\infty}$ $= - \frac{4}{25} \ln (2) + \frac{1}{5} + \frac{3 π}{25} = \frac{4}{25} \ln (\frac{1}{2}) + \frac{1}{5} + \frac{3 π}{25}$
	Commented by Ar Brandon last updated on 26/Jun/21
	$f(t)=(1/((t+2)^2 (t^2 +1)))=(a/(t+2))+(b/((t+2)^2 ))+((ct+d)/(t^2 +1)) =((a(t+2)(t^2 +1)+b(t^2 +1)+(ct+d)(t+2)^2 )/((t+2)^2 (t^2 +1))) lim_(t→−2) f(t)⇒5b=1⇒b=(1/5) lim_(t→i) f(t)⇒(3d−4c)+(3c+4d)i=1 ⇒ { ((3d−4c=1)),((4d+3c=0)) :}⇒d=(3/(25)), c=−(4/(25)) For coefficient of t^3 we have a+c=0⇒a=(4/(25)) ⇒f(t)=(4/(25))∙(1/(t+2))+(1/5)∙(1/((t+2)^2 ))−(1/(25))∙((4t−3)/(t^2 +1))$
	$f (t) = \frac{1}{{(t + 2)}^{2} (t^{2} + 1)} = \frac{a}{t + 2} + \frac{b}{{(t + 2)}^{2}} + \frac{ct + d}{t^{2} + 1}$ $= \frac{a (t + 2) (t^{2} + 1) + b (t^{2} + 1) + (ct + d) {(t + 2)}^{2}}{{(t + 2)}^{2} (t^{2} + 1)}$ $\underset{t \to - 2}{\lim f} (t) \Rightarrow 5 b = 1 \Rightarrow b = \frac{1}{5}$ $\underset{t \to i}{\lim f} (t) \Rightarrow (3 d - 4 c) + (3 c + 4 d) i = 1$ $\Rightarrow {\begin{cases} 3 d - 4 c = 1 \\ 4 d + 3 c = 0 \end{cases} \Rightarrow d = \frac{3}{25}, c = - \frac{4}{25}$ $For coefficient of t^{3} we have a + c = 0 \Rightarrow a = \frac{4}{25}$ $\Rightarrow f (t) = \frac{4}{25} \cdot \frac{1}{t + 2} + \frac{1}{5} \cdot \frac{1}{{(t + 2)}^{2}} - \frac{1}{25} \cdot \frac{4 t - 3}{t^{2} + 1}$
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