integrate 1/(sinx + cosx) dx

integrate 1/(sinx + cosx) dx
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∫ [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