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

The integral of (cos(2x-1))/(e^x) is

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integral of (cos(2x-1))/(e^x)

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

cos (1) (\frac{4}{5} e^{- x} cos (2 x) + \frac{2}{5} e^{- x} sin (2 x) - \frac{2 cos ^{2} ( x )}{e ^{x}} + \frac{1}{e ^{x}}) + (- \frac{2 e ^{- x} cos ( 2 x )}{5} - \frac{e ^{- x} sin ( 2 x )}{5}) sin (1) + C

Solution steps

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

Rewrite using trig identities

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

Expand

\frac{cos ( 2 x ) cos ( 1 ) + sin ( 2 x ) sin ( 1 )}{e ^{x}} : \frac{cos ( 1 ) cos ( 2 x )}{e ^{x}} + \frac{sin ( 1 ) sin ( 2 x )}{e ^{x}}

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int \frac{cos ( 1 ) cos ( 2 x )}{e ^{x}} d x = cos (1) (\frac{4}{5} e^{- x} cos (2 x) + \frac{2}{5} e^{- x} sin (2 x) - \frac{2 cos ^{2} ( x )}{e ^{x}} + \frac{1}{e ^{x}})

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

= cos (1) (\frac{4}{5} e^{- x} cos (2 x) + \frac{2}{5} e^{- x} sin (2 x) - \frac{2 cos ^{2} ( x )}{e ^{x}} + \frac{1}{e ^{x}}) + (- \frac{2 e ^{- x} cos ( 2 x )}{5} - \frac{e ^{- x} sin ( 2 x )}{5}) sin (1)

Add a constant to the solution

= cos (1) (\frac{4}{5} e^{- x} cos (2 x) + \frac{2}{5} e^{- x} sin (2 x) - \frac{2 cos ^{2} ( x )}{e ^{x}} + \frac{1}{e ^{x}}) + (- \frac{2 e ^{- x} cos ( 2 x )}{5} - \frac{e ^{- x} sin ( 2 x )}{5}) sin (1) + C

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What is the integral of (cos(2x-1))/(e^x) ?
The integral of (cos(2x-1))/(e^x) is