integrate 1/(sinx + cosx) dx

3 ANSWERS

1. paafamily
   

   ∫ [1 /(cosx + sinx)] dx =
   
   multiply and divide the integrand by (cosx - sinx):
   
   ∫ {(cosx - sinx) /[(cosx + sinx)(cosx - sinx)]} dx =
   
   expand the denominator:
   
   ∫ [(cosx - sinx) /(cos²x - sin²x)] dx =
   
   break it up into:
   
   ∫ [cosx /(cos²x - sin²x)] dx + ∫ [- sinx /(cos²x - sin²x)] dx =
   
   rewrite the first denominator in terms of sinx and the second one in terms of cosx:
   
   ∫ {cosx /[(1 - sin²x) - sin²x]} dx + ∫ {- sinx /[cos²x - (1 - cos²x)]} dx =
   
   ∫ [cosx /(1 - sin²x - sin²x)] dx + ∫ [- sinx /(cos²x - 1 + cos²x)] dx =
   
   ∫ [cosx /(1 - 2sin²x)] dx + ∫ [- sinx /(2cos²x - 1)] dx (#)
   
   let us solve the first integral substituting sinx = t hence (differentiating both sides)
   d(sinx) = dt → cosx dx = dt, yielding:
   
   ∫ cosx dx /(1 - 2sin²x) = ∫ dt /(1 - 2t²) =
   
   factor the denominator as a difference of squares:
   
   ∫ dt /{1 - [(√2)t]²} =
   
   ∫ dt /{[1 - (√2)t][1 + (√2)t]} =
   
   decompose it into partial fractions:
   
   1 /{[1 - (√2)t][1 + (√2)t]} = A/[1 - (√2)t] + B/[1 + (√2)t]
   
   1 /{[1 - (√2)t][1 + (√2)t]} = {A[1 + (√2)t] + B[1 - (√2)t]} /{[1 - (√2)t][1 + (√2)t]}
   
   1 = A + (√2)At + B - (√2)Bt
   
   1 = (√2)(A - B)t + (A + B)
   
   hence:
   
   | (√2)(A - B) = 0
   | A + B = 1
   
   | A = B
   | B + B = 1
   
   | A = 1/2
   | B = 1/2
   
   yielding:
   
   1 /{[1 - (√2)t][1 + (√2)t]} = A/[1 - (√2)t] + B/[1 + (√2)t] = (1/2)/[1 - (√2)t] + (1/2)/[1 + (√2)t]
   
   thus the integral becomes:
   
   ∫ [1 /(1 - 2t²)+ dt = ∫ {{(1/2)/[1 - (√2)t]} + {(1/2)/[1 + (√2)t]}} dt =
   
   break it up pulling constants out:
   
   (1/2) ∫ {1 /[1 - (√2)t]} dt + (1/2) ∫ {1 /[1 + (√2)t]} dt =
   
   divide and multiply the first integral by (-√2), and the second one by (√2) so as to make each numerator the derivative of the respective denominator:
   
   (1/2)(-1/√2) ∫ {(-√2) /[1 - (√2)t]} dt + (1/2)(1/√2) ∫ {(√2) /[1 + (√2)t]} dt =
   
   [- 1/(2√2)] ∫ d[1 - (√2)t] /[1 - (√2)t]} + [1/(2√2)] ∫ d[1 + (√2)t] /[1 + (√2)t] =
   
   [- 1/(2√2)] ln |1 - (√2)t| + [1/(2√2)] ln |1 + (√2)t| + C =
   
   [1/(2√2)] [ln |1 + (√2)t| - ln |1 - (√2)t|] + C =
   
   recalling logarithms properties,
   
   [1/(2√2)] ln |[1 + (√2)t] /[1 - (√2)t]| + C
   
   thus, substituiting back sinx for t, you have:
   
   ∫ [cosx /(1 - 2sin²x)] dx = [1/(2√2)] ln |[1 + (√2)sinx] /[1 - (√2)sinx]| + C (##)
   
   
   the solution of the latter integral (see expression (#) above) is similar:
   
   ∫ [- sinx /(2cos²x - 1)] dx =
   
   let cosx = u → d(cosx) = du → - sinx dx = du
   
   ∫ [- sinx /(2cos²x - 1)] dx = ∫ du /(2u² - 1) = ∫ du /[(√2)u]² - 1] =
   
   ∫ du /{[(√2)u - 1][(√2)u + 1]}
   
   partial fraction decomposition
   
   1 /{[(√2)u - 1][(√2)u + 1]} = A/[(√2)u - 1] + B/[(√2)u + 1]
   
   1 /{[(√2)u - 1][(√2)u + 1]} = {A[(√2)u + 1] + B[(√2)u - 1]} /{[(√2)u - 1][(√2)u + 1]}
   
   1 = A[(√2)u + 1] + B[(√2)u - 1]
   
   1 = (√2)Au + A + (√2)Bu - B
   
   1 = (√2)(A + B)u + (A - B)
   
   | (√2)(A + B) = 0
   | A - B = 1
   
   | A = - B
   | - B - B = 1
   
   | A = 1/2
   | B = - 1/2
   
   yielding:
   
   1 /{[(√2)u - 1][(√2)u + 1]} = A/[(√2)u - 1] + B/[(√2)u + 1] =
   
   (1/2)/[(√2)u - 1] - (1/2)/[(√2)u + 1]
   
   hence:
   
   ∫ du /(2u² - 1) = ∫ {{(1/2)/[(√2)u - 1]} - {(1/2)/[(√2)u + 1]}} du =
   
   (1/2) ∫ du /[(√2)u - 1] - (1/2) ∫ du /[(√2)u + 1] =
   
   dividing and multiplying by √2,
   
   (1/2)(1/√2) ∫ (√2) du /[(√2)u - 1] - (1/2)(1/√2) ∫ (√2) du /[(√2)u + 1] =
   
   [1/(2√2)] ∫ d[(√2)u - 1] /[(√2)u - 1] - [1/(2√2)] ∫ d[(√2)u + 1] /[(√2)u + 1] =
   
   [1/(2√2)] ln |(√2)u - 1| - [1/(2√2)] ln |(√2)u + 1| + C =
   
   [1/(2√2)] [ln |(√2)u - 1| - ln |(√2)u + 1|] + C =
   
   [1/(2√2)] ln |[(√2)u - 1] /[(√2)u + 1]| + C =
   
   substituiting back cosx for u,
   
   ∫ [- sinx /(2cos²x - 1)] dx = [1/(2√2)] ln |[(√2)cosx - 1] /[(√2)cosx + 1]| + C (###)
   
   thus, plugging this and the previous (###) result into the above (#) expression, you have:
   
   ∫ [cosx /(1 - 2sin²x)] dx + ∫ [- sinx /(2cos²x - 1)] dx = [1/(2√2)] ln |[1 + (√2)sinx] /[1 -
   
   (√2)sinx]| + [1/(2√2)] ln |[(√2)cosx - 1] /[(√2)cosx + 1]| + C =
   
   [1/(2√2)] {ln |[1 + (√2)sinx] /[1 - (√2)sinx]| + ln |[(√2)cosx - 1] /[(√2)cosx + 1]|} + C =
   
   owing to logarithm properties,
   
   [1/(2√2)] ln |{[1 + (√2)sinx] /[1 - (√2)sinx]}{[(√2)cosx - 1] /[(√2)cosx + 1]|} + C
   
   thus, in conclusion:
   
   ∫ [1 /(cosx + sinx)] dx =
   
   [1/(2√2)] ln |{[1 + (√2)sinx][(√2)cosx - 1]} /{[1 - (√2)sinx][(√2)cosx + 1]}| + C

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2. Guest9547
   

   ∫▒cos⁡〖x□(24&dx)〗/√(1+sin⁡x )

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3. Guest3377
   

    I = integration of 1+sinx/1+cosx dx

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