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Question Number 118340 by bramlexs22 last updated on 17/Oct/20

   ∫_(π/3) ^(π/2)  (dx/(1+sin x−cos x))

π/2π/3dx1+sinxcosx

Commented by Dwaipayan Shikari last updated on 17/Oct/20

∫_(π/3) ^(π/2) (dx/(1−cosx+sinx))=2∫_(1/( (√3))) ^1 (dt/(1−((1−t^2 )/(1+t^2 ))+((2t)/(1+t^2 )))).(1/(1+t^2 ))       x=tan(x/2)  =2∫_(1/( (√3))) ^1 (dt/(2t^2 +2t))=∫_(1/( (√3))) ^1 (dt/(t(t+1)))=[log((t/(t+1)))]_(1/( (√3))) ^1   =log((1/2))−log((1/( (√3)+1)))=log((((√3)+1)/2))

π3π2dx1cosx+sinx=2131dt11t21+t2+2t1+t2.11+t2x=tanx2=2131dt2t2+2t=131dtt(t+1)=[log(tt+1)]131=log(12)log(13+1)=log(3+12)

Answered by Bird last updated on 17/Oct/20

I =∫_(π/3) ^(π/2)   (dx/(1+sinx−cosx))  we do the ch.tan((x/2))=t ⇒  I =∫_(1/( (√3))) ^1    ((2dt)/((1+t^2 )(1+((2t)/(1+t^2 ))−((1−t^2 )/(1+t^2 )))))  =∫_(1/( (√3))) ^1   ((2dt)/(1+t^2 +2t−1+t^2 ))  =∫_(1/( (√3))) ^1   ((2dt)/(2t^2 +2t)) =∫_(1/( (√3))) ^1  (dt/(t(t+1)))  =∫_(1/( (√3))) ^1  ((1/t)−(1/(t+1)))dt  =[ln∣(t/(t+1))∣]_(1/( (√3))) ^1 =ln((1/2))−ln((1/( (√3)(1+(1/( (√3)))))))  =−ln(2)+ln(1+(√3))

I=π3π2dx1+sinxcosxwedothech.tan(x2)=tI=1312dt(1+t2)(1+2t1+t21t21+t2)=1312dt1+t2+2t1+t2=1312dt2t2+2t=131dtt(t+1)=131(1t1t+1)dt=[lntt+1]131=ln(12)ln(13(1+13))=ln(2)+ln(1+3)

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