Question

integration of e^-x cosx​

Answer 1

We have to find,

$\displaystyle\int e^{-x}\cos x\ dx$

By product rule of integration, let,

$u'=e^{-x}\quad\implies\quad u=-e^{-x}$

How?

$\displaystyle u=\int e^{-x}\ dx\\\\\\u=\int e^{-x}\cdot-d(-x)\quad \left[\because\ \dfrac {d(-x)}{dx}=-1\right]\\\\\\u=-\int e^{-x}\ d(-x)\\\\\\u=-\dfrac {e^{-x}}{\ln(e)}\quad \left[\because\ \int\ a^x\ dx=\dfrac {a^x}{\ln(a)}+c\right]\\\\\\u=-e^{-x}$

And,

$v=\cos x\quad\implies\quad v'=-\sin x$

Thus, by product rule,

$\displaystyle\int u'v=uv-\int uv'\\\\\\\int e^{-x}\cos x\ dx=-e^{-x}\cos x-\int-e^{-x}\cdot-\sin x\ dx\\\\\\\int e^{-x}\cos x\ dx=-e^{-x}\cos x-\int e^{-x}\cdot\sin x\ dx$

Now, consider the integral,

$\displaystyle\int e^{-x}\cdot\sin x\ dx$

By product rule,

$\displaystyle\int e^{-x}\cdot\sin x\ dx=-e^{-x}\sin x-\int-e^{-x}\cos x$

Then,

$\displaystyle\int e^{-x}\cos x\ dx=-e^{-x}\cos x-\left (-e^{-x}\sin x-\int-e^{-x}\cos x\ dx\right)\\\\\\\int e^{-x}\cos x\ dx=-e^{-x}\cos x-\left (-e^{-x}\sin x+\int e^{-x}\cos x\ dx\right)\\\\\\\int e^{-x}\cos x\ dx=-e^{-x}\cos x+e^{-x}\sin x-\int e^{-x}\cos x\ dx\\\\\\2\int e^{-x}\cos x\ dx=-e^{-x}\cos x+e^{-x}\sin x\\\\\\\boxed {\boxed {\int e^{-x}\cos x\ dx=\dfrac {e^{-x}(\sin x-\cos x)}{2}+c}}$

Hence integrated!

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