Question Number 179137 by yaslm last updated on 25/Oct/22

Answered by MJS_new last updated on 26/Oct/22

∫((sin 3x)/( (√(1−sin x cos x))))dx= [t=tan x → dx=(dt/(t^2 +1))] =−∫((t(t^2 −3))/((t^2 +1)^2 (√(t^2 −t+1))))dt= [u=(((√3)(2t−1))/3) → dt=((√3)/2)du] =−2∫((3(√3)u^3 +9u^2 −9(√3)u−11)/( (√(u^2 +1))(3u^2 +2(√3)u+5)^2 ))du= [v=u+(√(u^2 +1)) → du=((√(u^2 +1))/v)dv] =−4∫((3(√3)v^6 +18v^5 −45(√3)v^4 −124v^3 +45(√3)v^2 +18v−3(√3))/((3v^4 +4(√3)v^3 +14v^2 −4(√3)v+3)^2 ))dv= [w=v+((√3)/3) → dv=dw] =−((4(√3))/3)∫((w^6 −20w^4 +((32(√3))/3)w^3 +((64)/3)w^2 −((128(√3))/9)w+((128)/(27)))/((w^2 −((2(√6))/3)w+((4(2−(√2)))/3))^2 (w^2 +((2(√6))/3)w+((4(2+(√2)))/3))^2 ))dw= [Ostrogradski′s Method] =((((4(√3))/3)w^3 +((16)/3)w^2 −((80(√3))/9)w+((32)/3))/((w^2 −((2(√6))/3)w+((4(2−(√2)))/3))(w^2 +((2(√6))/3)w+((4(2+(√2)))/3))))+((32)/3)∫((w−((√3)/3))/((w^2 −((2(√6))/3)w+((4(2−(√2)))/3))(w^2 +((2(√6))/3)w+((4(2+(√2)))/3))))dw= [decompose & formula] =((((4(√3))/3)w^3 +((16)/3)w^2 −((80(√3))/9)w+((32)/3))/((w^2 −((2(√6))/3)w+((4(2−(√2)))/3))(w^2 +((2(√6))/3)w+((4(2+(√2)))/3))))+ +((√2)/2)ln ((3w^2 −2(√6)w+4(2−(√2)))/(3w^2 +2(√6)w+4(2+(√2)))) + +(√2)arctan ((2(√2)((√3)w−2))/(3w^2 )) that′s the path, I might have made some mistakes though, it′s a bit complicated...

Commented byMJS_new last updated on 26/Oct/22

the answer is ∫_0 ^(π/2) ((sin 3x)/( (√(1−sin x cos x))))dx=2((√2)ln (1+(√2)) −1)

Commented byyaslm last updated on 26/Oct/22

great sir . thank you