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

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

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

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

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

Solution steps

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

Apply u-substitution

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

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

Apply Integration By Parts

: - \frac{1}{4} e^{u} cos (4 u) - \int - \frac{1}{4} e^{u} cos (4 u) d u

= 2 (- \frac{1}{4} e^{u} cos (4 u) - \int - \frac{1}{4} e^{u} cos (4 u) d u)

Take the constant out:

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

= 2 (- \frac{1}{4} e^{u} cos (4 u) - (- \frac{1}{4} \cdot \int e^{u} cos (4 u) d u))

Apply Integration By Parts

: e^{u} \frac{1}{4} sin (4 u) - \int e^{u} \frac{1}{4} sin (4 u) d u

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

Take the constant out:

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

= 2 (- \frac{1}{4} e^{u} cos (4 u) - (- \frac{1}{4} (e^{u} \frac{1}{4} sin (4 u) - \frac{1}{4} \cdot \int e^{u} sin (4 u) d u)))

Therefore

2 \cdot \int e^{u} sin (4 u) d u = 2 (- \frac{1}{4} e^{u} cos (4 u) - (- \frac{1}{4} (e^{u} \frac{1}{4} sin (4 u) - \frac{1}{4} \cdot \int e^{u} sin (4 u) d u)))

Isolate

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

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

Substitute back

u = \frac{x}{2}

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

Simplify

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

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

Add a constant to the solution

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

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