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Calculate-pi-6-0-cos-2-x-1-2sinx-dx-

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Question Number 138703 by mathocean1 last updated on 16/Apr/21
Calculate ∫_(−(π/6)) ^0 ((cos^2 x)/(1−2sinx))dx
Calculate∫0−π6cos2x1−2sinxdx
Commented by SanyamJoshi last updated on 17/Apr/21
Calculate ∫_(−(π/6)) ^0 ((cos^2 x)/(1−2sinx))dx
Calculate∫0−π6cos2x1−2sinxdx
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=∫−π60cos2x1−2sinxdx=x=−t∫0π6cos2t1+2sintdt=∫0π61−sin2t1+2sintdtwehave1−x22x+1=−144x2−42x+1=−14(4x2−12x+1−32x+1)=−14(2x−1)+34(2x+1)⇒I=−14∫0π6(2sint−1)dt+34∫0π6dt2sint+1(=J)=−12[−cost]0π6+14.π6+34J=−12(1−32)+π24+34JJ=∫0π6dt2sint+1=tan(t2)=y∫02−32dy(1+y2)(22y1+y2+1)=2∫02−3dy4y+y2+1=2∫02−3dyy2+4y+1Δ′=22−1=3⇒y1=−2+3y2=−2−3⇒2∫02−3dy(y−y1)(y−y2)=2123∫02−3(1y−y1−1y−y2)dy=13[log∣y+2−3y+2+3∣]02−3=13{log(4−234)−log(2−32+3)}I=−12+34+π24+34{log(1−32)−log(2−32+3)}

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