Math, asked by hpharry666, 1 year ago

intregration
of sin x Cos 2x by 1 + sin x

Answers

Answered by sabanaz20032001
4

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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