intregration
of sin x Cos 2x by 1 + sin x
Answers
Answer:
Step-by-step explanation:
I=sin^2 (x) /(1+sinx cosx)
=[{(1-cos2x)/2}/(1+sinx cosx)]
=[(1-cos2x)/(2+2sinxcosx)]
=[(1-cos2x)/(2+sin2x)]
=[1/(2+sin2x) -cos2x/(2+sin2x)]
=[1/(2+2tanx/(1+tan^2 x) -cos2x/(2+sin2x)]
=[sec^2 x /(2tan^2 x +2tanx+2)-cos2x/(2+sin2x)]
Put tanx=t and 2+sin2x=z
Differentiate w.r.t.x
sec^2 x dx=dt and cos2x dx=dx/2
I=[dt/2(t^2 +t+1) -1/2{dx/z}]
=1/2[dt/{(t+1/2)^2 +3/4}-dz/z]
=1/2[dt/{(t+1/2)^2 +(√3/2)^2}-dz/z]
Integrate
I=1/2[2/√3arctan{(t+1/2)/√3/2} -ln|z|] +c
=1/√3 arctan2tanx+1/ √3 -1/2 ln|2+sin2x| +c
552 Views · View 2 Upvoters · Answer requested by Gourav Saini
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Arkajyoti Banerjee
Arkajyoti Banerjee, Math nerd, Codebug, Trap & Death Metal is my type of music.
Answered Mar 23
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