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

The integral of e^{-x/2}sin(2x) is -(8e^{-x/2}cos(2x))/(17)-(2e^{-x/2}sin(2x))/(17)+C

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

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

- \frac{8 e ^{- \frac{x}{2}} cos ( 2 x )}{17} - \frac{2 e ^{- \frac{x}{2}} sin ( 2 x )}{17} + C

Solution steps

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

Apply Integration By Parts

: - \frac{1}{2} e^{- \frac{x}{2}} cos (2 x) - \int \frac{1}{4} e^{- \frac{x}{2}} cos (2 x) d x

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

Take the constant out:

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

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

Apply Integration By Parts

: \frac{1}{2} e^{- \frac{x}{2}} sin (2 x) - \int - \frac{1}{4} e^{- \frac{x}{2}} sin (2 x) d x

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

Take the constant out:

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

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

Therefore

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

Isolate

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

= - \frac{8 e ^{- \frac{x}{2}} cos ( 2 x )}{17} - \frac{2 e ^{- \frac{x}{2}} sin ( 2 x )}{17}

Add a constant to the solution

= - \frac{8 e ^{- \frac{x}{2}} cos ( 2 x )}{17} - \frac{2 e ^{- \frac{x}{2}} sin ( 2 x )}{17} + C

x \to 0 + lim ((1 + x)^{l n (x)})

\int x arccos (x) d x

\frac{1}{s ^{2} + 2 s + 3}

\int_{2}^{9} \frac{1}{7 ( x - 2 ) ^{3}} d x

x \to - 7 - lim (\frac{1}{x ^{2} ( x + 7 )})

What is the integral of e^{-x/2}sin(2x) ?
The integral of e^{-x/2}sin(2x) is -(8e^{-x/2}cos(2x))/(17)-(2e^{-x/2}sin(2x))/(17)+C