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bemath-lim-x-pi-1-2x-2pi-pi-x-cos-2t-dt-1-cos-3t-




Question Number 106869 by bemath last updated on 07/Aug/20
               @bemath@  lim_(x→π)  [((1/(2x−2π))) ∫_π ^x  ((cos 2t dt)/(1−cos 3t)) ]=?
@bemath@limxπ[(12x2π)xπcos2tdt1cos3t]=?
Answered by bobhans last updated on 07/Aug/20
          ^(∧bobhans∧)   lim_(x→π)  [((∫_π ^x  ((cos 2t dt)/(1−cos 3t)))/(2x−2π)) ] = lim_(x→π) (((d/dx) ∫^x  ((cos 2t dt)/(1−cos 3t)))/((d/dx)[2x−2π]))  = lim_(x→π)  (((((cos 2x)/(1−cos 3x))))/2) = ((1/(1+1))/2) = (1/4)
bobhanslimxπ[xπcos2tdt1cos3t2x2π]=limxπddxxcos2tdt1cos3tddx[2x2π]=limxπ(cos2x1cos3x)2=11+12=14

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