how to do
notice，My brackets
∫ 1/((2 + cosx) sinx) dx
Let t = tan(x/2)
dt = 1/2 sec²(x/2) dx
2 cos²(x/2) dt = dx
From tan(x/2) = t, you have
sin(x/2) = t/√(1 + t²)
cos(x/2) = 1/√(1 + t²)
Squaring both sides, you have
cos²(x/2) = 1/(1 + t²), so the equation containing dt and dx becomes
2/(1 + t²) dt = dx
cosx = cos(2 (x/2)) = cos²(x/2) - sin²(x/2)
cosx = [1/√(1 + t²)]² - [t/√(1 + t²)]²
cosx = (1 - t²)/(1 + t²)
sinx = sin(2 (x/2)) = 2 sin(x/2) cos(x/2)
sinx = 2 (t/√(1 + t²)) (1/√(1 + t²))
sinx = 2t/(1 + t²)
The integral then changes from
to
∫ 1/((2 + (1 - t²)/(1 + t²)) (2t/(1 + t²))) ∙ (2/(1 + t²) dt)
∫ [2/(1 + t²)] / [ (2 + (1 - t²)/(1 + t²)) ∙ 2t/(1 + t²) ] dt
∫ 1 / [ (2 + (1 - t²)/(1 + t²)) ∙ t ] dt
∫ 1 / [ (2 + 2t² + 1 - t²)/(1 + t²)) ∙ t ] dt
∫ 1 / [ (3 + t²)/(1 + t²)) ∙ t ] dt
∫ (1 + t²) / [ (3 + t²)t ] dt
Break up the integrand into partial fractions:
(1 + t²) / (t (3 + t²)) = A/t + (Bt + D)/(3 + t²)
1 + t² = A(3 + t²) + (Bt + D)t
1 + t² = 3A + At² + Bt² + Dt
1 + t² = (A + B)t² + Dt + 3A
Matching up the coefficients, you have
A + B = 1
D = 0
3A = 1
From the third equation, you have A = 1/3.
In the first equation, you get B = 2/3.
(1 + t²) / (t (3 + t²)) = (1/3)/t + (2/3)t/(3 + t²)
1/3 ∫ 1/t dt + 2/3 ∫ t/(3 + t²) dt
In the second integral, let u = 3 + t²
du = 2t dt
1/2 du = t dt
1/3 ∫ 1/t dt + 2/3 (1/2) ∫ 1/u du
1/3 ln|t| + 1/3 ln|u| + C
1/3 ln|t| + 1/3 ln|3 + t²| + C
1/3 ln|tan(x/2)| + 1/3 ln|3 + tan²(x/2)| + C
2+cosx d/dx = -sinx
let
2+cosx =u
du = -sinx dx
then,
remember ∫du/u = ln u
yet, you need to multiply 1/(2+cosx) by -1/-1 to satisfy the fromula above
∫(-1/-1)[1/(2+cosx)](sinx) dx
then extract the constant 1/-1
1/-1 ∫1/(2+cosx) * -sinx dx
-1∫du/u = -ln u
where u is 2+cosx
∫1/[(2+cosx)*sinx] dx = -ln (2+cosx) +c
∫1/[(2+cosx)*sinx] dx =
∫sin(x) /[(2+cos(x))sin^2(x)] dx =
- ∫ 1/[(2+cos(x))(1-cos^2(x)] d(cos(x))
cos(x) = u
- ∫ 1/[(2 + u)(1-u^2)] du =
A∫1/(2+u)du +B∫1/(1+u) du + C∫1/(1-u) du
A(1-u^2) + B(1-u)(2+u) + C(2+u)(1+u) = - 1
C = -1/6
B = -1/2
A = 1/3
(1/3)∫1/(2+u)du - (1/2)∫1/(1+u) du -(1/6) ∫1/(1-u) du =
(1/3)Ln|2+u| - (1/2)Ln|1+u| -(1/6)Ln|1-u| + C Back to x
(1/3)Ln|2+cos(x)| - (1/2)Ln|1+cos(x)| -(1/6)Ln|1-cos(x)| + C
Comments
∫ 1/((2 + cosx) sinx) dx
Let t = tan(x/2)
dt = 1/2 sec²(x/2) dx
2 cos²(x/2) dt = dx
From tan(x/2) = t, you have
sin(x/2) = t/√(1 + t²)
cos(x/2) = 1/√(1 + t²)
Squaring both sides, you have
cos²(x/2) = 1/(1 + t²), so the equation containing dt and dx becomes
2/(1 + t²) dt = dx
cosx = cos(2 (x/2)) = cos²(x/2) - sin²(x/2)
cosx = [1/√(1 + t²)]² - [t/√(1 + t²)]²
cosx = (1 - t²)/(1 + t²)
sinx = sin(2 (x/2)) = 2 sin(x/2) cos(x/2)
sinx = 2 (t/√(1 + t²)) (1/√(1 + t²))
sinx = 2t/(1 + t²)
The integral then changes from
∫ 1/((2 + cosx) sinx) dx
to
∫ 1/((2 + (1 - t²)/(1 + t²)) (2t/(1 + t²))) ∙ (2/(1 + t²) dt)
∫ [2/(1 + t²)] / [ (2 + (1 - t²)/(1 + t²)) ∙ 2t/(1 + t²) ] dt
∫ 1 / [ (2 + (1 - t²)/(1 + t²)) ∙ t ] dt
∫ 1 / [ (2 + 2t² + 1 - t²)/(1 + t²)) ∙ t ] dt
∫ 1 / [ (3 + t²)/(1 + t²)) ∙ t ] dt
∫ (1 + t²) / [ (3 + t²)t ] dt
Break up the integrand into partial fractions:
(1 + t²) / (t (3 + t²)) = A/t + (Bt + D)/(3 + t²)
1 + t² = A(3 + t²) + (Bt + D)t
1 + t² = 3A + At² + Bt² + Dt
1 + t² = (A + B)t² + Dt + 3A
Matching up the coefficients, you have
A + B = 1
D = 0
3A = 1
From the third equation, you have A = 1/3.
In the first equation, you get B = 2/3.
(1 + t²) / (t (3 + t²)) = (1/3)/t + (2/3)t/(3 + t²)
1/3 ∫ 1/t dt + 2/3 ∫ t/(3 + t²) dt
In the second integral, let u = 3 + t²
du = 2t dt
1/2 du = t dt
1/3 ∫ 1/t dt + 2/3 (1/2) ∫ 1/u du
1/3 ln|t| + 1/3 ln|u| + C
1/3 ln|t| + 1/3 ln|3 + t²| + C
1/3 ln|tan(x/2)| + 1/3 ln|3 + tan²(x/2)| + C
2+cosx d/dx = -sinx
let
2+cosx =u
du = -sinx dx
then,
remember ∫du/u = ln u
yet, you need to multiply 1/(2+cosx) by -1/-1 to satisfy the fromula above
∫(-1/-1)[1/(2+cosx)](sinx) dx
then extract the constant 1/-1
1/-1 ∫1/(2+cosx) * -sinx dx
-1∫du/u = -ln u
where u is 2+cosx
∫1/[(2+cosx)*sinx] dx = -ln (2+cosx) +c
∫1/[(2+cosx)*sinx] dx =
∫sin(x) /[(2+cos(x))sin^2(x)] dx =
- ∫ 1/[(2+cos(x))(1-cos^2(x)] d(cos(x))
cos(x) = u
- ∫ 1/[(2 + u)(1-u^2)] du =
A∫1/(2+u)du +B∫1/(1+u) du + C∫1/(1-u) du
A(1-u^2) + B(1-u)(2+u) + C(2+u)(1+u) = - 1
C = -1/6
B = -1/2
A = 1/3
(1/3)∫1/(2+u)du - (1/2)∫1/(1+u) du -(1/6) ∫1/(1-u) du =
(1/3)Ln|2+u| - (1/2)Ln|1+u| -(1/6)Ln|1-u| + C Back to x
(1/3)Ln|2+cos(x)| - (1/2)Ln|1+cos(x)| -(1/6)Ln|1-cos(x)| + C