What is the integral of e^{x/2}sin(3x) ?

The integral of e^{x/2}sin(3x) is 2(-6/37 e^{x/2}cos(3x)+1/37 e^{x/2}sin(3x))+C

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integral of e^{x/2}sin(3x)

\int e^{\frac{x}{2}} sin (3 x) d x

2 (- \frac{6}{37} e^{\frac{x}{2}} cos (3 x) + \frac{1}{37} e^{\frac{x}{2}} sin (3 x)) + C

Solution steps

\int e^{\frac{x}{2}} sin (3 x) d x

Apply u-substitution

: \int 2 e^{u} sin (6 u) d u

= \int 2 e^{u} sin (6 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 (6 u) d u

Apply Integration By Parts

: - \frac{1}{6} e^{u} cos (6 u) - \int - \frac{1}{6} e^{u} cos (6 u) d u

= 2 (- \frac{1}{6} e^{u} cos (6 u) - \int - \frac{1}{6} e^{u} cos (6 u) d u)

Take the constant out:

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

= 2 (- \frac{1}{6} e^{u} cos (6 u) - (- \frac{1}{6} \cdot \int e^{u} cos (6 u) d u))

Apply Integration By Parts

: e^{u} \frac{1}{6} sin (6 u) - \int e^{u} \frac{1}{6} sin (6 u) d u

= 2 (- \frac{1}{6} e^{u} cos (6 u) - (- \frac{1}{6} (e^{u} \frac{1}{6} sin (6 u) - \int e^{u} \frac{1}{6} sin (6 u) d u)))

Take the constant out:

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

= 2 (- \frac{1}{6} e^{u} cos (6 u) - (- \frac{1}{6} (e^{u} \frac{1}{6} sin (6 u) - \frac{1}{6} \cdot \int e^{u} sin (6 u) d u)))

Therefore

2 \cdot \int e^{u} sin (6 u) d u = 2 (- \frac{1}{6} e^{u} cos (6 u) - (- \frac{1}{6} (e^{u} \frac{1}{6} sin (6 u) - \frac{1}{6} \cdot \int e^{u} sin (6 u) d u)))

Isolate

\int e^{u} sin (6 u) d u

= 2 (- \frac{6 e ^{u} cos ( 6 u )}{37} + \frac{e ^{u} sin ( 6 u )}{37})

Substitute back

u = \frac{x}{2}

= 2 (- \frac{6 e ^{\frac{x}{2}} cos ( 6 \cdot \frac{x}{2} )}{37} + \frac{e ^{\frac{x}{2}} sin ( 6 \cdot \frac{x}{2} )}{37})

Simplify

2 (- \frac{6 e ^{\frac{x}{2}} cos ( 6 \cdot \frac{x}{2} )}{37} + \frac{e ^{\frac{x}{2}} sin ( 6 \cdot \frac{x}{2} )}{37}) : 2 (- \frac{6}{37} e^{\frac{x}{2}} cos (3 x) + \frac{1}{37} e^{\frac{x}{2}} sin (3 x))

= 2 (- \frac{6}{37} e^{\frac{x}{2}} cos (3 x) + \frac{1}{37} e^{\frac{x}{2}} sin (3 x))

Add a constant to the solution

= 2 (- \frac{6}{37} e^{\frac{x}{2}} cos (3 x) + \frac{1}{37} e^{\frac{x}{2}} sin (3 x)) + C

\frac{\partial}{\partial y} (3 x - 2 y^{4})

\int \frac{x ^{2}}{x ^{3} - 2} d x

x = - 2, x = 3, y = x^{2} - 12, y = x - 6

1 - x^{12}

What is the integral of e^{x/2}sin(3x) ?
The integral of e^{x/2}sin(3x) is 2(-6/37 e^{x/2}cos(3x)+1/37 e^{x/2}sin(3x))+C