Calculate ∫_(−(π/6)) ^0 ((cos^2 x)/(1−2sinx))dx


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Question Number 138703 by mathocean1 last updated on 16/Apr/21

$C a l c u l a t e$ $\underset{- \frac{π}{6}}{\int^{0}} \frac{{c o s}^{2} x}{1 - 2 s i n x} d x$
Commented by SanyamJoshi last updated on 17/Apr/21

$C a l c u l a t e$ $\underset{- \frac{π}{6}}{\int^{0}} \frac{{c o s}^{2} x}{1 - 2 s i n x} d x$
	Answered by Mathspace last updated on 16/Apr/21
	$I=∫_(−(π/6)) ^0 ((cos^2 x)/(1−2sinx))dx =_(x=−t) ∫_0 ^(π/6) ((cos^2 t)/(1+2sint))dt =∫_0 ^(π/6) ((1−sin^2 t)/(1+2sint))dt we have ((1−x^2 )/(2x+1))=−(1/4)((4x^2 −4)/(2x+1)) =−(1/4)(((4x^2 −1)/(2x+1))−(3/(2x+1))) =−(1/4)(2x−1)+(3/(4(2x+1))) ⇒ I=−(1/4)∫_0 ^(π/6) (2sint−1)dt +(3/4)∫_0 ^(π/6) (dt/(2sint +1))(=J) =−(1/2)[−cost]_0 ^(π/6) +(1/4).(π/6) +(3/4)J =−(1/2)(1−((√3)/2))+(π/(24))+(3/4)J J=∫_0 ^(π/6) (dt/(2sint +1)) =_(tan((t/2))=y) ∫_0 ^(2−(√3)) ((2dy)/((1+y^2 )(2((2y)/(1+y^2 ))+1))) =2∫_0 ^(2−(√3)) (dy/(4y+y^2 +1)) =2∫_0 ^(2−(√3)) (dy/(y^2 +4y+1)) Δ^′ =2^2 −1=3 ⇒y_1 =−2+(√3) y_2 =−2−(√3) ⇒ 2∫_0 ^(2−(√3)) (dy/((y−y_1 )(y−y_2 ))) =2(1/(2(√3)))∫_0 ^(2−(√3)) ((1/(y−y_1 ))−(1/(y−y_2 )))dy =(1/( (√3)))[log∣((y+2−(√3))/(y+2+(√3)))∣]_0 ^(2−(√3)) =(1/( (√3))){log(((4−2(√3))/4))−log(((2−(√3))/(2+(√3))))} I=−(1/2)+((√3)/4)+(π/(24))+((√3)/4){log(1−((√3)/2))−log(((2−(√3))/(2+(√3))))}$
	$I = \int_{- \frac{π}{6}}^{0} \frac{{c o s}^{2} x}{1 - 2 s i n x} d x$ $=_{x = - t} \int_{0}^{\frac{π}{6}} \frac{{c o s}^{2} t}{1 + 2 s i n t} d t$ $= \int_{0}^{\frac{π}{6}} \frac{1 - {s i n}^{2} t}{1 + 2 s i n t} d t$ $w e h a v e \frac{1 - x^{2}}{2 x + 1} = - \frac{1}{4} \frac{4 x^{2} - 4}{2 x + 1}$ $= - \frac{1}{4} (\frac{4 x^{2} - 1}{2 x + 1} - \frac{3}{2 x + 1})$ $= - \frac{1}{4} (2 x - 1) + \frac{3}{4 (2 x + 1)} \Rightarrow$ $I = - \frac{1}{4} \int_{0}^{\frac{π}{6}} (2 s i n t - 1) d t$ $+ \frac{3}{4} \int_{0}^{\frac{π}{6}} \frac{d t}{2 s i n t + 1} (= J)$ $= - \frac{1}{2} {[- c o s t]}_{0}^{\frac{π}{6}} + \frac{1}{4} . \frac{π}{6}$ $+ \frac{3}{4} J$ $= - \frac{1}{2} (1 - \frac{\sqrt{3}}{2}) + \frac{π}{24} + \frac{3}{4} J$ $J = \int_{0}^{\frac{π}{6}} \frac{d t}{2 s i n t + 1}$ $=_{t a n (\frac{t}{2}) = y} \int_{0}^{2 - \sqrt{3}} \frac{2 d y}{(1 + y^{2}) (2 \frac{2 y}{1 + y^{2}} + 1)}$ $= 2 \int_{0}^{2 - \sqrt{3}} \frac{d y}{4 y + y^{2} + 1}$ $= 2 \int_{0}^{2 - \sqrt{3}} \frac{d y}{y^{2} + 4 y + 1}$ $Δ^{^{'}} = 2^{2} - 1 = 3 \Rightarrow y_{1} = - 2 + \sqrt{3}$ $y_{2} = - 2 - \sqrt{3} \Rightarrow$ $2 \int_{0}^{2 - \sqrt{3}} \frac{d y}{(y - y_{1}) (y - y_{2})}$ $= 2 \frac{1}{2 \sqrt{3}} \int_{0}^{2 - \sqrt{3}} (\frac{1}{y - y_{1}} - \frac{1}{y - y_{2}}) d y$ $= \frac{1}{\sqrt{3}} {[l o g ∣ \frac{y + 2 - \sqrt{3}}{y + 2 + \sqrt{3}} ∣]}_{0}^{2 - \sqrt{3}}$ $= \frac{1}{\sqrt{3}} {l o g (\frac{4 - 2 \sqrt{3}}{4}) - l o g (\frac{2 - \sqrt{3}}{2 + \sqrt{3}})}$ $I = - \frac{1}{2} + \frac{\sqrt{3}}{4} + \frac{π}{24} + \frac{\sqrt{3}}{4} {l o g (1 - \frac{\sqrt{3}}{2}) - l o g (\frac{2 - \sqrt{3}}{2 + \sqrt{3}})}$
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