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Question Number 145791 by mathmax by abdo last updated on 08/Jul/21

find lim_(x→0)    ∫_0 ^x  ((e^t +e^(−t) −2)/(1−cosx))dx

findlimx00xet+et21cosxdx

Commented by mathmax by abdo last updated on 09/Jul/21

lim_(x→0)   ∫_0 ^x  ((e^t +e^(−t) −2)/(1−cosx))dt

limx00xet+et21cosxdt

Commented by mathmax by abdo last updated on 10/Jul/21

let use hospital rule  =lim_(x→0)    (1/(1−cosx))(e^x  +e^(−x) −2) =lim_(x→0)    ((e^x −e^(−x) )/(sinx))  =lim_(x→0)    (((1+x+(x^2 /2))−(1−x +(x^2 /2)))/(sinx))=lim_(x→0)    ((2x)/(sinx))=2

letusehospitalrule=limx011cosx(ex+ex2)=limx0exexsinx=limx0(1+x+x22)(1x+x22)sinx=limx02xsinx=2

Answered by Olaf_Thorendsen last updated on 09/Jul/21

f(x) = ∫_0 ^x ((e^t +e^(−t) −2)/(1−cosx)) dx  f(x) = [((2sht−2t)/(1−cosx))]_0 ^x   f(x) = ((2shx−2x)/(1−cosx))  f(x) ∼_0  ((2(x+(x^3 /(3!)))−2x)/(1−(1−(x^2 /(2!))))) ∼_0  (x/3) → 0

f(x)=0xet+et21cosxdxf(x)=[2sht2t1cosx]0xf(x)=2shx2x1cosxf(x)02(x+x33!)2x1(1x22!)0x30

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