Question

найти определенный интеграл

Answer 1

a
t=√(2x-3)+7,dt=dx/√(2x-3)
$intlimits^6_2 {(4x+1)/( sqrt{2x-3} +7)} , dx = intlimits {(2t^3+7t)/(t+7)} , dt=$
2t³+7t      |t+7
2t³+14t²    2t²-14t+105
------------
     -14t²+7t
     -14t²-98t
   ---------------
           105y
           105t+735
         ---------------
                 - 735
=$intlimits {(2t^2-14t+105-735/(t+7)} , dt =2/3* sqrt{(2x-3)^3} -7t^2+105t-$$735ln( sqrt{2x-3} +7)|6-2=2/3*1-7*9+105*3-735ln10-2/3*1$$+7*1-105*1+735ln8=$161 1/3+735ln0,8
b
sin(2x-π/4)*cosx=sin2x*cosπ/4*cosx-cos2x*sinπ/4*cosx=
=√2/2*sin2xcosx-√2/2*cos2xcosx=√2/2*1/2*(sinx+sin3x-cosx-cos3x)=
=√2/4*(sinx+sin3x-cosx-cos3x)
$intlimits^{ pi /2}_{- pi /2} {sin(2x- pi /4)cosx} , dx =$$sqrt{2} /4* intlimits^{ pi /2}_{- pi /2} {(sinx+sin3x-cosx-cos3x)} , dx =$$sqrt{2} /4(-cosx-1/3*cos3x-sinx-1/3*sin3x)| pi /2-(- pi /2)=$$sqrt{2}/4*(-0-1/3*0-1+1/3*1 +0+1/3*0+1-1/3*1)=$$sqrt{2} /4*(-2)=- sqrt{2} /2$