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

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

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integral of (cos(x))/(e^{2x)}

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

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

Solution steps

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

Apply exponent rule:

\frac{1}{a ^{b}} = a^{- b}

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

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

Apply Integration By Parts

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

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

Apply Integration By Parts

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

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

Therefore

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

Isolate

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

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

Add a constant to the solution

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

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