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raheen

  • 4 years ago

solve this integration in details using integration by parts:(−e−x)cosxdx

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  1. raheen
    • 4 years ago
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    \[\int\limits (-e ^{-x})cosx dx\]

  2. imranmeah91
    • 4 years ago
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    \[-\int\limits (e ^{-x})cosx dx\]

  3. jahtoday
    • 4 years ago
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    \[\int\limits_{}^{}udv=uv-\int\limits_{}^{}vdu\] u=cos(x) \[dv=-e^{-x}\]

  4. jahtoday
    • 4 years ago
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    so: \[\int\limits_{}^{}-e ^{-x}\cos(x)dx=e ^{-x} \cos(x)-\int\limits_{}^{}-e^{-x}\sin(x)\]

  5. jahtoday
    • 4 years ago
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    and u=sin(x) dv=-e^(-x) so: \[\int\limits_{}^{}-e^{-x}\cos(x)dx=e^{-x}\cos(x)-e^{-x}\sin(x)+\int\limits_{}^{}e^{-x}\cos(x)\]

  6. jahtoday
    • 4 years ago
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    subtract \[\int\limits_{}^{} e^{-x}\cos(x) \] from both sides: \[-2\int\limits_{}^{}e^{-x}\cos(x)=e^{-x}\cos(x)-e^{-x}\sin(x)\] divide each side by -2: \[\int\limits\limits_{}^{}e^{-x}\cos(x)=(-1/2)(e^{-x}\cos(x)-e^{-x}\sin(x))+C\]

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