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Question Number 118340 by bramlexs22 last updated on 17/Oct/20
∫π/2π/3dx1+sinx−cosx
Commented by Dwaipayan Shikari last updated on 17/Oct/20
∫π3π2dx1−cosx+sinx=2∫131dt1−1−t21+t2+2t1+t2.11+t2x=tanx2=2∫131dt2t2+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=∫π3π2dx1+sinx−cosxwedothech.tan(x2)=t⇒I=∫1312dt(1+t2)(1+2t1+t2−1−t21+t2)=∫1312dt1+t2+2t−1+t2=∫1312dt2t2+2t=∫131dtt(t+1)=∫131(1t−1t+1)dt=[ln∣tt+1∣]131=ln(12)−ln(13(1+13))=−ln(2)+ln(1+3)
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