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Question Number 183279 by cortano1 last updated on 24/Dec/22

∫ ((sin x+cos x)/((tan x−cot x)^3 )) dx =?

∫sinx+cosx(tanx−cotx)3dx=?

Answered by MJS_new last updated on 24/Dec/22

with t=tan (x/2) we get −16∫((t^3 (t^2 −1)^3 )/((t^2 +1)^2 (t^2 −2t−1)^2 (t^2 +2t−1)^3 ))dt= [Ostrogradski′s Method] =−(((t+1)(9t^6 −57t^4 −104t^3 +51t^2 +24t−11))/(32(t^2 +1)(t^2 −2t−1)(t^2 +2t−1)^2 ))−(9/(32))∫(dt/(t^2 +2t−1))= =−(((t+1)(9t^6 −57t^4 −104t^3 +51t^2 +24t−11))/(32(t^2 +1)(t^2 −2t−1)(t^2 +2t−1)^2 ))+((9(√2))/(128))ln ((t+1+(√2))/(t+1−(√2))) ...

witht=tanx2weget−16∫t3(t2−1)3(t2+1)2(t2−2t−1)2(t2+2t−1)3dt=[Ostrogradski′sMethod]=−(t+1)(9t6−57t4−104t3+51t2+24t−11)32(t2+1)(t2−2t−1)(t2+2t−1)2−932∫dtt2+2t−1==−(t+1)(9t6−57t4−104t3+51t2+24t−11)32(t2+1)(t2−2t−1)(t2+2t−1)2+92128lnt+1+2t+1−2...

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