| seif | Date: Tuesday, 11/Jun/2013, 10:44 | Message # 1 |
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| IntegrateCosxSinx dx use three different methods. I only know one method  .... can anyone help me about other topics?
Yes I can
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| Admin | Date: Tuesday, 11/Jun/2013, 10:46 | Message # 2 |
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| First Method:
Int, CosxSinx dx let U = Sinx ------> du/dx = Cosx then dx = du/Cosx
Int, Cosx U (du/Cosx) hapo Cosxitaondoka na Cosx iliyopo ndani ya Bracket
Int, U du
= ½ U2 + C C is constant.
Since U = Sin x then:
=1/2Sin2x +C
Second Method:
Int, CosxSinx dx let U = Cosx ------> du/dx = -Sinx then dx = du/-Sinx
Int, U Sinx (du/-Sinx) hapo Sinx itaondoka na Sinx iliyopo ndani ya Bracket.)
Int, -U du
= -½ U2 + C C is constant.
Since U = Cos x then:
=-1/2 Cos2x +C NOTE: Be careful withNegative Sign
Third Method:
Remember that sin2x= 2SinxCosx so by dividing by 2 bothsides we obtain1/2Sin2x = SinxCosx
Therefore: CosxSinx= 1/2Sin2x then integrate now 1/2Sin2x. this is the third method:
Int, 1/2Sin2x let U = 2x ------> du/dx = 2 then dx = du/2
Int, 1/2SinU X du/2
Int, 1/4SinU du
= -1/4CosU + C But U = 2x
Therefore:
=-1/4Cos2x +C
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