Question Number 114072 by Lordose last updated on 17/Sep/20

∫(1/(sinx + cosx))dx

Answered by Olaf last updated on 17/Sep/20

x = (π/4)−u  sinx+cosx =   (sin(π/4)cosu−sinucos(π/4))+(cos(π/4)cosu+sin(π/4)sinu)  = ((√2)/2)(cosu−sinu)+((√2)/2)(cosu+sinu)  = (√2)cosu  −(1/( (√2)))∫(du/(cosu)) = −(1/( (√2)))∫(du/((1−t^2 )/(1+t^2 )))  with t = tan(u/2)  dt = (1/2)(1+tan^2 (u/2))du  du = ((2dt)/(1+t^2 ))  −(1/( (√2)))∫((1+t^2 )/(1−t^2 )).((2dt)/(1+t^2 ))  = −(√2)∫(dt/(1−t^2 )) = −(1/( (√2)))∫((1/(1−t))+(1/(1+t)))dt  = (1/( (√2)))ln∣((1−t)/(1+t))∣ = (1/( (√2)))ln∣((1−tan(u/2))/(1+tan(u/2)))∣  (1/( (√2)))ln∣((1−tan((π/8)−(x/2)))/(1+tan((π/8)−(x/2))))∣...

Answered by MJS_new last updated on 17/Sep/20

∫(dx/(sin x +cos x))=       [t=x+(π/4) → dx=dt]  =((√2)/2)∫csc t dt=((√2)/2)ln ∣tan (t/2)∣=  =((√2)/2)ln ∣tan ((x/2)+(π/8))∣=  =((√2)/2)ln ∣(((√2)+2sin x)/( (√2)+2cos x))∣ +C

Answered by Dwaipayan Shikari last updated on 17/Sep/20

∫(1/(sinx+cosx))dx  =(√2)∫(1/(cos((π/4)−x)))dx  =(√2) ∫sec((π/4)−x)dx  =−(√2) ∫secudu=−(√2) log(secu+tanu)  =−(√2)  log(sec((π/4)−x)+tan((π/4)−x))

Commented byMJS_new last updated on 17/Sep/20

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