Question

integration of 1/1+2cosx dx​

Answer 1

Answer:

∫ [1 /(1 + 2cosx)] dx =

let:

tan(x/2) = t

x/2 = arctan t

x = 2arctan t

dx = 2 [1/(1 + t²)] dt

also, recall the tangent half-angle formula:

cosx = [1 - tan²(x/2)]/[1 + tan²(x/2)] = (1 - t²)/(1 + t²)

then, substituting:

∫ [1 /(1 + 2cosx)] dx = ∫ {1 /{1 + 2[(1 - t²) /(1 + t²)]} } 2 [1 /(1 + t²)] dt =

∫ {1 /{[(1 + t²) + 2(1 - t²)] /(1 + t²)} } 2 [1 /(1 + t²)] dt =

∫ {1 /[(1 + t² + 2 - 2t²) /(1 + t²)]} 2 [1 /(1 + t²)] dt =

∫ {1 /[(3 - t²) /(1 + t²)]} 2 [1 /(1 + t²)] dt =

∫ [(1 + t²) /(3 - t²)] 2 [1 /(1 + t²)] dt =

simplifying into:

∫ [2 /(3 - t²)] dt =

let's factor the denominator (as a difference of squares):

∫ {2 /[(√3)² - t²]} dt =

∫ {2 /{[(√3) + t][(√3) - t]} } dt =

let's decompose the integrand into partial fractions:

2 /{[(√3) + t][(√3) - t]} = A/[(√3) + t] + B/[(√3) - t]

2 /{[(√3) + t][(√3) - t]} = {A[(√3) - t] + B[(√3) + t]} /{[(√3) + t][(√3) - t]}

(equating numerators)

2 = A[(√3) - t] + B[(√3) + t]

2 = (√3)A - At + (√3)B + Bt

2 = (- A + B)t + (√3)(A + B)

- A + B = 0

(√3)(A + B) = 2

B = A

(√3)(A + A) = 2

B = A

(√3)2A = 2

B = A = 1/√3

A = 1/√3

therefore:

2 /{[(√3) + t][(√3) - t]} = A/[(√3) + t] + B/[(√3) - t] = (1/√3)/[(√3) + t] + (1/√3)/[(√3) - t]

thus the integral becomes:

∫ [2 /(3 - t²)] dt = ∫ { {(1/√3)/[(√3) + t]} + {(1/√3)/[(√3) - t]} } dt =

let's split this pulling constants out:

(1/√3) ∫ {1 /[(√3) + t]} dt + (1/√3) ∫ {1 /[(√3) - t]} dt =

(changing the sign of the second integral to make the top the derivative of the bottom)

(1/√3) ln |(√3) + t| - (1/√3) ∫ [1 /(t - √3)] dt =

(1/√3) ln |t + √3| - (1/√3) ln |t - √3| + C =

(1/√3) (ln |t + √3| - ln |t - √3|) + C =

(recalling logarithm properties)

(1/√3) ln |(t + √3)/(t - √3)| + C

substitute back tan(x/2) for t, ending with:

∫ [1 /(1 + 2cosx)] dx = (1/√3) ln |[tan(x/2) + √3]/[tan(x/2) - √3]| + C

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