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1/(5+4cosx)^2 0^pi

1/(5+4cosx)^2 0^pi

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1 Answers

Rinkoo Gupta
askIITians Faculty 81 Points
10 years ago
integral dx/(5+4cosx)^2
put cosx=(1-tan^2(x/2))/(1+tan^2(x/2))
integral sec^4(x/2) dx/(9+tan^2(x/2))^2
put tan(x/2)=t
diff.w.r.to x we get
(sec^2(x/2)) dx= 2dt
integral 2(1+t^2)dt/(9+t^2)^2
=2integral{(t^2+9)-8}/(9+t^2)^2
=2integral[1/(t^2+9) -8/(9+t^2)^2]
=2[1/3 arctan(t/3)-8 integral dt/(9+t^2)^2]
=2/3 arctan(t/3)-16 integral dt/(9+t^2)^2
put t=3tan@
=>dt=3sec^2@d@
=2/3 arctan(t/3)-16integral 3sec^2@d@/81sec^4@
=2/3arctan(t/3)-16/27integral cos^2@d@
=2/3arctan(t/3)-16/54integral (1+cos2@)d@
=2/3arctan(t/3)-8/27(@+1/2sin2@)
=2/3arctan(1/3 .tan(x/2))-8/27 {arctan(t/3)+(3tan(x/2))/(9+tan^2(x/2))}
=2/3arctan(1/3tan(x/2))-8/27{arctan(1/3.tan(x/2))+(3tan(x/2))/(9+tan^2(x/2))
=10/27.arctan(1/3tan(x/2))-8/9 (tan(x/2))/(9+tan^2(x/2))

Hence integral( lim 0 to pi) dx/(5+4cosx)^2
=2integral lim 0 to pi/2 dx/(5+4cosx)^2
hence the value of the integral is
2[10/27arctan(1/3tan(x/2) -8/9 .(tan(x/2))/(9+tan^2(x/2)) lim 0 to pi/2
=2[10/27arctan(1/3tan(pi/4) -8/9 (tan(pi/4))/(9+tan^2(pi/4))]
=2[10/27arctan(1/3)-(8/9)(1/10)]
=2[10/27arctan(1/3)-4/45]
Thanks & Regards
Rinkoo Gupta
AskIITians Faculty

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