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Question Number 71814 by Aman Arya last updated on 20/Oct/19
∫11+cotxdx
Commented by mathmax by abdo last updated on 20/Oct/19
letI=∫dx1+cotanx⇒I=∫dx1+cosxsinx=∫sinxsinx+cosxdxchangementtan(x2)=tgiveI=∫2t1+t22t1+t2+1−t21+t22dt1+t2I=∫2t(1+t2)2{2t1+t2+1−t21+t2}dt=∫2t(1+t2)(−t2+2t+1)dt=−2∫t(t2+1)(t2−2t−1)dtletdecomposeF(t)=t(t2+1)(t2−2t−1)t2−2t−1=0→Δ′=1+1=2⇒t1=1+2andt2=1−2F(t)=t(t−t1)(t−t2)(t2+1)=at−t1+bt−t2+ct+dt2+1a=limt→t1(t−t1)F(t)=t1(t1−t2)(t12+1)=1+222(3+22+1)=2+24(4+22)=2+28(2+2)=18b=t2(t2−t1)(t22+1)=1−2(−22)(3−22+1)=−1+222(4−22)=−2+28(2−2)=18limt→+∞tF(t)=0=a+b+c⇒c=−a−b=−14⇒F(t)=18(t−t1)+18(t−t2)+−14t+dt2+1F(0)=0=−18t1−18t2+d⇒d=18(1t1+1t2)=18(t1+t2t1t2)=182−1=−14⇒F(t)=18(t−t1)+18(t−t2)−14t+1t2+1⇒I=−14∫dtt−t1−14∫dtt−t2+12∫t+1t2+1dt=−14ln∣t−t1∣−14ln∣t−t2∣+14ln(t2+1)+12arctant+c=−14ln∣tan(x2)−1−2∣−14ln∣tan(x2)−1+2∣+14ln(1+tan2(x2))+x4+c.
Commented by petrochengula last updated on 21/Oct/19
=∫11+cosxsinxdx=∫sinxsinx+cosxdx=12∫2sinxsinx+cosxdx=12∫sinx+cosxsinx+cosxdx−12∫cosx−sinxsinx+cosxdx=12x−12ln∣sinx+cosx∣+C
Answered by $@ty@m123 last updated on 20/Oct/19
∫11+cosxsinxdx∫sinxsinx+cosxdx∫12(sinx+cosx)−12(cosx−sinx)sinx+cosxdx=12∫dx−12∫d(sinx+cosx)sinx+cosx=12x−12ln(sinx+cosx)+C
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