Question Number 101212 by rb222 last updated on 01/Jul/20

∫e^x sin x dx = −e^x cos x + e^x  sin x − ∫ e^x  sin  x dx

Commented byDwaipayan Shikari last updated on 01/Jul/20

∫e^x sinxdx=e^x ∫sinxdx+∫e^x cosxdx         I=−e^x cosx+e^x sinx−∫e^x sinxdx  I=e^x (sinx−cosx)−I  2I=e^x (sinx−cosx)+C  I=((e^x (sinx−cosx))/2)+Constant(or (C/2))

Answered by smridha last updated on 01/Jul/20

Im∫e^((1+i)x) dx=Im[(e^((1+i)x) /((1+i)))]+c  =(1/2)Im[(1−i)e^x (cosx+isinx)]+c  =(1/2)Im[e^x {(cosx+sinx)+i(sinx−cosx)}]+c  =(e^x /2)(sinx−cosx)+c     ans  or =(e^x /(√2))(sinxcos(𝛑/4)−cosx.sin(𝛑/4))+c     =(e^x /(√2))sin(x−(𝛑/4))+c        ans  or =(e^x /(√2))cos[(𝛑/2)+(x−(𝛑/4))]+c      =(e^x /(√2))cos[x+(𝛑/4)]+c        ans  let     I=∫e^x sinx=e^x sinx−∫e^x cosx             =e^x sinx−e^x cosx−∫e^x sinx             =e^x (sinx−cosx)−I+k  2I=e^x (sinx−cosx)+k  so I=(e^x /2)(sinx−cosx)+c    [c=(k/2)=constant]

Commented bysmridha last updated on 01/Jul/20

who are you man ??well copied!!

Answered by MJS last updated on 01/Jul/20

why so complicated?  it′s simply 2 times by parts  I=∫e^x sin x dx=       u′=u=e^x        v=sin x → v′=cos x  =e^x sin x −∫e^x cos x dx=       u′=u=e^x        v=cos x → v′=−sin x  =e^x sin x −(e^x cos x −∫−e^x sin x dx)=  =e^x sin x −e^x cos x −∫e^x sin x dx  now we have  I=e^x sin x −e^x cos x −I  2I=e^x sin x −e^x cos x  I=((e^x sin x −e^x cos x)/2)+C

Commented bysmridha last updated on 02/Jul/20

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