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| Pregunta | Respuesta |
| Reverse differentition Integration | d/dx sinx=cosx so integrate cosxdx= sinx+c d/dx cosx=-sinx so integrate sinxdx= -cosx+c |
| Trig Integrals | integrate 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 |
| Example 1- Integrate (sec(2)3x)dx Example 2- Integrate tan(2x-1/4pie)dx Example 3- Integrate (x(2)-sin3x)dx | E.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 |
| Double angle identities formula | Integrate 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 |
| Example 4- Integrate tan(2)x dx Example 5- integrate sin(2) dx Example 6- integrate cos(2)x dx | E.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 |
| Use identities sin(A+B) and sin(A-B) to show that sin(A+B) +sin(A-B)= 2sinAsinB | 2sinAsinB=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 |
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