Найти определенный интеграл

Question 1

Question 2

A
t=√(2x-3)+7,dt=dx/√(2x-3)
$\int\limits^6_2 {(4x+1)/( \sqrt{2x-3} +7)} \, dx = \int\limits {(2t^3+7t)/(t+7)} \, dt=$
2t³+7t    |t+7
2t³+14t² 2t²-14t+105
------------
   -14t²+7t
   -14t²-98t
   ---------------
   105y
   105t+735
   ---------------
   - 735
= $\int\limits {(2t^2-14t+105-735/(t+7)} \, dt =2/3* \sqrt{(2x-3)^3} -7t^2+105t-$ $735ln( \sqrt{2x-3} +7)|6-2=2/3*1-7*9+105*3-735ln10-2/3*1$ $+7*1-105*1+735ln8=$ 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)
$\int\limits^{ \pi /2}_{- \pi /2} {sin(2x- \pi /4)cosx} \, dx =$ $\sqrt{2} /4* \int\limits^{ \pi /2}_{- \pi /2} {(sinx+sin3x-cosx-cos3x)} \, dx =$ $\sqrt{2} /4(-cosx-1/3*cos3x-sinx-1/3*sin3x)| \pi /2-(- \pi /2)=$ $\sqrt{2}/4*(-0-1/3*0-1+1/3*1 +0+1/3*0+1-1/3*1)=$ $\sqrt{2} /4*(-2)=- \sqrt{2} /2$

sedinalana_zn · Answer 1 · 2018-05-07T20:26:02+0000

A
t=√(2x-3)+7,dt=dx/√(2x-3)
$\int\limits^6_2 {(4x+1)/( \sqrt{2x-3} +7)} \, dx = \int\limits {(2t^3+7t)/(t+7)} \, dt=$
2t³+7t    |t+7
2t³+14t² 2t²-14t+105
------------
   -14t²+7t
   -14t²-98t
   ---------------
   105y
   105t+735
   ---------------
   - 735
= $\int\limits {(2t^2-14t+105-735/(t+7)} \, dt =2/3* \sqrt{(2x-3)^3} -7t^2+105t-$ $735ln( \sqrt{2x-3} +7)|6-2=2/3*1-7*9+105*3-735ln10-2/3*1$ $+7*1-105*1+735ln8=$ 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)
$\int\limits^{ \pi /2}_{- \pi /2} {sin(2x- \pi /4)cosx} \, dx =$ $\sqrt{2} /4* \int\limits^{ \pi /2}_{- \pi /2} {(sinx+sin3x-cosx-cos3x)} \, dx =$ $\sqrt{2} /4(-cosx-1/3*cos3x-sinx-1/3*sin3x)| \pi /2-(- \pi /2)=$ $\sqrt{2}/4*(-0-1/3*0-1+1/3*1 +0+1/3*0+1-1/3*1)=$ $\sqrt{2} /4*(-2)=- \sqrt{2} /2$