Flashcards
Mathematics-C4
Flashcards do conjunto (6)
- 1Reverse differentition Integrationd/dx sinx=cosx so integrate cosxdx= sinx+c
d/dx cosx=-sinx so integrate sinxdx= -cosx+c - 2Trig Integralsintegrate tanxdx=In[secx]+k
integrate sec2xdx=tanx+k
integrate secxtanxdx= secx+k
integrate cotxdx=In[sinx]+k
integrate cosecxdx=-In[cosex+cotx]+k
integrate secxdx=In[secx+tanx]+k - 3Example 1- Integrate (sec(2)3x)dx Example 2- Integrate tan(2x-1/4pie)dx Example 3- Integrate (x(2)-sin3x)dxE.g 1) sec(2)x=tanx+k
=1/3tan3x+k
E.g 2) tanx=In[secx]+k
(2x-1/4pie)=1/2(2x-1/4pie)
=1/2In[sec(2x-1/4pie)]+k
E.g 3) -sinx=cosx+k
=1/3x(3)+1/3cos(3)x+k - 4Double angle identities formulaIntegrate sin(2)x= 1/2sin2x+1/2
Integrate cos(2)x= 1/2cos2x+1/2
Integrate cot(2)x= cosec(2)x-1 dx
Integrate tan(2)x= sec(2)x-1 dx - 5Example 4- Integrate tan(2)x dx Example 5- integrate sin(2) dx Example 6- integrate cos(2)x dxE.g 4) tan(2)x=sec(2)x-1dx
sec(2)x=tanx+k
=tanx-x+k
E.g 5) sin(2)x=1/2sin2x+1/2
sin2x=-cos2x+k
=-1/4cos2x+1/2x+k
E.g 6) cos(2)=1/2cos2x+1/2
cos2x=sin2x+k
=1/4sin2x+1/2x+k - 6Use identities sin(A+B) and sin(A-B) to show that sin(A+B) +sin(A-B)= 2sinAsinB2sinAsinB=sinAsinB+sinAsinB
sin2A=2sinAsinA
SO: 1sinAsinA=1/2sin2A
Hence: sin5xcos3x=1/2{sin(5x+3x)+sin(5x-3x)}
=1/2sin8x+1/2sin2x
Therefore integrate 1/2sin8x+sin2xdx
=-(1/16cos8x+1/4cos2x)+c