3/(1+2cos x) dx [20 points!!]

∫3/(1+2Cos x) dx
Sol
Set y=Tan(x/2)
dy=(1/2)Sec^2 (x/2)dx=(1/2)*(1+y^2)dx
dx=2/(1+y^2)dy
Cosx
=[Cos^2 (x/2)－Sin^2 (x/2)/[Cos^2(x/2)+Sin^2 (x/2)]
=(1－y^2)/(1+y^2)
1+2Cosx
=1+(2－2y^2)/(1+y^2)
=(3－y^2)/(1+y^2)
A=∫3/(1+2Cos x) dx
=∫3[(1+y^2)/(3－y^2)]*2/(1+y^2)dy
=6∫1/(3－y^2) dy
Set y=√3Sinw
dy=√3Coswdw
3－y^2=3－3Sin^2w=3Cos^2 w
A=6∫1/(3Cos^2 w)*√3Coswdw
=2√3∫1/Coswdw
=2√3∫Secwdw
=2√3ln(｜Secw+Tanw｜)+c
=2√3ln(｜√3/√(3－y^2)+y/√(3－y^2)｜)+c
=2√3ln(｜√3/√(3－Tan^2 (x/2))+y/√(3－Tan^2 (x/2)｜+c

Integration of the form 1/(a cos x + b sin x + c) dx, where a, b, c are constants, may be solved by using the Weierstrass substitution, t= tan (x/2).Consider a right-angled triangle ABC, angle B = 90 deg, Angle A = x/2, BC = t, AB = 1Hypotenuse AC = √(1+ t^2)tan (x/2) = t/1t = tan (x/2) ------- equation (1)sin (x/2) = t/√(1+ t^2)cos (x/2) = 1/ √(1+ t^2)Since sin x = 2 sin(x/2) cos (x/2) = 2[t/√(1+ t^2)][ 1/ √(1 + t^2)]sin x = 2t/(1 + t^2) ---------- equation (2)cos x = cos^2 (x/2) – sin^2 (x/2) = [1/ √ (1 + t^2)]^2 –[t/√ (1+ t^2)]^2cos x = (1 + t^2)/(1 + t^2) ------------ equation (3)Also t = tan (x/2)dt/dx = (1/2)sec^2 (x/2) =(1/2)( 1 + tan^2(x/2)dt/dx = (1/2)(1 + t^2)From which dx = 2 dt/(1 + t^2) ---------- equation (4) ∫ 3/(1+2cos x) dxSince t = tan (x/2), cos x = (1 + t^2)/(1 + t^2), dx = 2dt/(1 + t^2) 2 dt/(1+ t^2) 3 ∫ -------------------------------------- 1 + 2(1 +t^2)/(1 + t^2), Simplify dt6 ∫ ---------------- equation (5) 3 - t^2 dt6 ∫ --------------------------- [√3]^2- t^2 Using the formula ∫ 1/(a^2 – u^2) du =(1/2a) ln |(u – a)/(u + a)| + C where C is a constant ( t + √3)6 [1/(2 √3)] ln | ------------------ | + C ( t – √3) since t = tan (x/2), (tan (x/2) + √3)3/ √(3)ln | --------------------------- | + C (tan (x/2),– √3) √3 ln|[tan (x/2) + √3]/[ tan(x/2)– √3 ]|+C (ANSWER) [tan (x/2) +1.732]/[tan(x/2)– 1.732] must be > 0 because you are taking the natural log of itIt implies x/2 > 60 deg
Comment:The answer depends on what form you get in equation (5) after simplification.For example,∫ 1/(5 + 4 cos x) dx = (2/3) tan^(-1) [1/3 tan(x/2)] + C∫ 1/(3 +5 cos x) dx = (1/4) ln |(2 + tan(x/2)) / (2 -tan(x/2))| + C.

2013-04-23 04:02:50 補充：
Correction:

x/2 > 60 deg is not required.
I overlook the absolute sign
Absolute value of sny real number is positive