What is the integral of e^{1-x}cos(2x) ?

The integral of e^{1-x}cos(2x) is (2e^{1-x}sin(2x))/5-(e^{1-x}cos(2x))/5+C

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integral of e^{1-x}cos(2x)

\int e^{1 - x} cos (2 x) d x

\frac{2 e ^{1 - x} sin ( 2 x )}{5} - \frac{e ^{1 - x} cos ( 2 x )}{5} + C

Solution steps

\int e^{1 - x} cos (2 x) d x

Apply Integration By Parts

= \frac{1}{2} e^{1 - x} sin (2 x) - \int - \frac{1}{2} e^{1 - x} sin (2 x) d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} e^{1 - x} sin (2 x) - (- \frac{1}{2} \cdot \int e^{1 - x} sin (2 x) d x)

Apply Integration By Parts

= \frac{1}{2} e^{1 - x} sin (2 x) - (- \frac{1}{2} (- \frac{1}{2} e^{1 - x} cos (2 x) - \int \frac{1}{2} e^{1 - x} cos (2 x) d x))

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} e^{1 - x} sin (2 x) - (- \frac{1}{2} (- \frac{1}{2} e^{1 - x} cos (2 x) - \frac{1}{2} \cdot \int e^{1 - x} cos (2 x) d x))

Therefore

\int e^{1 - x} cos (2 x) d x = \frac{1}{2} e^{1 - x} sin (2 x) - (- \frac{1}{2} (- \frac{1}{2} e^{1 - x} cos (2 x) - \frac{1}{2} \cdot \int e^{1 - x} cos (2 x) d x))

Isolate

\int e^{1 - x} cos (2 x) d x

= \frac{2 e ^{1 - x} sin ( 2 x )}{5} - \frac{e ^{1 - x} cos ( 2 x )}{5}

Add a constant to the solution

= \frac{2 e ^{1 - x} sin ( 2 x )}{5} - \frac{e ^{1 - x} cos ( 2 x )}{5} + C

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What is the integral of e^{1-x}cos(2x) ?
The integral of e^{1-x}cos(2x) is (2e^{1-x}sin(2x))/5-(e^{1-x}cos(2x))/5+C