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

The integral of e^{-2x}sin(x/2) is 2(-1/17 e^{-2x}cos(x/2)-4/17 e^{-2x}sin(x/2))+C

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integral of e^{-2x}sin(x/2)

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

2 (- \frac{1}{17} e^{- 2 x} cos (\frac{x}{2}) - \frac{4}{17} e^{- 2 x} sin (\frac{x}{2})) + C

Solution steps

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

Apply u-substitution

: \int 2 e^{- 4 u} sin (u) d u

= \int 2 e^{- 4 u} sin (u) d u

Take the constant out:

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

= 2 \cdot \int e^{- 4 u} sin (u) d u

Apply Integration By Parts

: - e^{- 4 u} cos (u) - \int 4 e^{- 4 u} cos (u) d u

= 2 (- e^{- 4 u} cos (u) - \int 4 e^{- 4 u} cos (u) d u)

Take the constant out:

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

= 2 (- e^{- 4 u} cos (u) - 4 \cdot \int e^{- 4 u} cos (u) d u)

Apply Integration By Parts

: e^{- 4 u} sin (u) - \int - 4 e^{- 4 u} sin (u) d u

= 2 (- e^{- 4 u} cos (u) - 4 (e^{- 4 u} sin (u) - \int - 4 e^{- 4 u} sin (u) d u))

Take the constant out:

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

= 2 (- e^{- 4 u} cos (u) - 4 (e^{- 4 u} sin (u) - (- 4 \cdot \int e^{- 4 u} sin (u) d u)))

Therefore

2 \cdot \int e^{- 4 u} sin (u) d u = 2 (- e^{- 4 u} cos (u) - 4 (e^{- 4 u} sin (u) - (- 4 \cdot \int e^{- 4 u} sin (u) d u)))

Isolate

\int e^{- 4 u} sin (u) d u

= 2 (- \frac{e ^{- 4 u} cos ( u )}{17} - \frac{4 e ^{- 4 u} sin ( u )}{17})

Substitute back

u = \frac{x}{2}

= 2 (- \frac{e ^{- 4 \cdot \frac{x}{2}} cos ( \frac{x}{2} )}{17} - \frac{4 e ^{- 4 \cdot \frac{x}{2}} sin ( \frac{x}{2} )}{17})

Simplify

2 (- \frac{e ^{- 4 \cdot \frac{x}{2}} cos ( \frac{x}{2} )}{17} - \frac{4 e ^{- 4 \cdot \frac{x}{2}} sin ( \frac{x}{2} )}{17}) : 2 (- \frac{1}{17} e^{- 2 x} cos (\frac{x}{2}) - \frac{4}{17} e^{- 2 x} sin (\frac{x}{2}))

= 2 (- \frac{1}{17} e^{- 2 x} cos (\frac{x}{2}) - \frac{4}{17} e^{- 2 x} sin (\frac{x}{2}))

Add a constant to the solution

= 2 (- \frac{1}{17} e^{- 2 x} cos (\frac{x}{2}) - \frac{4}{17} e^{- 2 x} sin (\frac{x}{2})) + C

f (x) = arctan (\frac{2}{x})

\int x \cdot cos (k x) d x

- 10 x^{4} + x^{3}

y = x^{\frac{13}{14}}

n = 1 \sum \infty ln ((n + 1)^{2})

What is the integral of e^{-2x}sin(x/2) ?
The integral of e^{-2x}sin(x/2) is 2(-1/17 e^{-2x}cos(x/2)-4/17 e^{-2x}sin(x/2))+C