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## Main Question or Discussion Point

I want to integrate [itex] \frac{e^x}{cos(x)} [/itex] with eulers formula.

I start by writing [itex] \frac{e^x}{e^{ix}} [/itex]

then I integrate that as usual.

So after I integrate I get [itex] \frac{e^{(1-i)x}}{1-i} [/itex]

Normally I would multiply and divide by the complex conjugate and then back substitute in

e^(ix) and the take the real part.

can I just back substitute in isin(x)+cos(x) on the bottom and then multiply it by (1-i)

and then take the real part. That seems to easy.

Does anyone have suggestions. This is not a homework problem.

I start by writing [itex] \frac{e^x}{e^{ix}} [/itex]

then I integrate that as usual.

So after I integrate I get [itex] \frac{e^{(1-i)x}}{1-i} [/itex]

Normally I would multiply and divide by the complex conjugate and then back substitute in

e^(ix) and the take the real part.

can I just back substitute in isin(x)+cos(x) on the bottom and then multiply it by (1-i)

and then take the real part. That seems to easy.

Does anyone have suggestions. This is not a homework problem.