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

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

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

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

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

Solution steps

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

Apply Integration By Parts

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

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

Take the constant out:

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

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

Apply Integration By Parts

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

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

Take the constant out:

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

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

Therefore

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

Isolate

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

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

Add a constant to the solution

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

y^{^{'}} = y^{2} - 2

y = (4 x - x^{2})^{3}

\frac{d}{d x} (\frac{4 x}{( x ^{2} + 1 ) ^{2}})

\int_{- 3}^{4} x d x

f (x) = 4 x^{2} - 5 x - 4 x - 2

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