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Question Number 48104 by wasim last updated on 19/Nov/18
solve this ∫(2 sinx+cosx)/(2+3sinx+sin^(2x) ) dx
solvethis∫(2sinx+cosx)/(2+3sinx+sin2x)dx
Answered by MJS last updated on 19/Nov/18
Weierstrass−substitution t=tan (x/2) ⇒ x=2arctan t; dx=((2dt)/(1+t^2 )) sin x =((2t)/(1+t^2 )); cos x =((1−t^2 )/(1+t^2 )) −∫((t^2 −4t−1)/((t+1)^2 (t^2 +t+1)))dt= =2∫(dt/(t+1))−4∫(dt/((t+1)^2 ))−∫((2t−3)/(t^2 +t+1))dt= =2∫(dt/(t+1))−4∫(dt/((t+1)^2 ))−∫((2x+1)/(t^2 +t+1))dt+4∫(dt/(t^2 +t+1)) =2ln (t+1) +(4/(t+1))−ln (t^2 +t+1) +((8(√3))/3)arctan (((√3)/3)(2t+1)) = =ln (((t+1)^2 )/(t^2 +t+1)) +(4/(t+1))+((8(√3))/3)arctan (((√3)/3)(2t+1)) = =ln ((1+sin x)/(2+sin x)) +ln 2 +2+2sec x −2tan x +((8(√3))/3)arctan (((√3)/3)(1+2tan (x/2))) = =ln ((1+sin x)/(2+sin x)) +2sec x −2tan x +((8(√3))/3)arctan (((√3)/3)(1+2tan (x/2))) +C
Weierstrass−substitutiont=tanx2⇒x=2arctant;dx=2dt1+t2sinx=2t1+t2;cosx=1−t21+t2−∫t2−4t−1(t+1)2(t2+t+1)dt==2∫dtt+1−4∫dt(t+1)2−∫2t−3t2+t+1dt==2∫dtt+1−4∫dt(t+1)2−∫2x+1t2+t+1dt+4∫dtt2+t+1=2ln(t+1)+4t+1−ln(t2+t+1)+833arctan(33(2t+1))==ln(t+1)2t2+t+1+4t+1+833arctan(33(2t+1))==ln1+sinx2+sinx+ln2+2+2secx−2tanx+833arctan(33(1+2tanx2))==ln1+sinx2+sinx+2secx−2tanx+833arctan(33(1+2tanx2))+C

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