What is the integral of e^{yt}*sin(kt) ?

The integral of e^{yt}*sin(kt) is -(ke^{yt}cos(kt))/(k^2+y^2)+(ye^{yt}sin(kt))/(k^2+y^2)+C

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integral of e^{yt}*sin(kt)

\int e^{y t} \cdot sin (k t) d t

- \frac{k e ^{y t} cos ( k t )}{k ^{2} + y ^{2}} + \frac{y e ^{y t} sin ( k t )}{k ^{2} + y ^{2}} + C

Solution steps

\int e^{y t} sin (k t) d t

Apply Integration By Parts

= - e^{y t} \frac{1}{k} cos (k t) - \int - y e^{y t} \frac{1}{k} cos (k t) d t

Take the constant out:

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

= - e^{y t} \frac{1}{k} cos (k t) - (- y \frac{1}{k} \cdot \int e^{y t} cos (k t) d t)

Simplify

- y \frac{1}{k} : - \frac{y}{k}

= - e^{y t} \frac{1}{k} cos (k t) - (- \frac{y}{k} \cdot \int e^{y t} cos (k t) d t)

Apply Integration By Parts

= - e^{y t} \frac{1}{k} cos (k t) - (- \frac{y}{k} (e^{y t} \frac{1}{k} sin (k t) - \int e^{y t} y \frac{1}{k} sin (k t) d t))

Take the constant out:

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

= - e^{y t} \frac{1}{k} cos (k t) - (- \frac{y}{k} (e^{y t} \frac{1}{k} sin (k t) - y \frac{1}{k} \cdot \int e^{y t} sin (k t) d t))

Simplify

y \frac{1}{k} : \frac{y}{k}

= - e^{y t} \frac{1}{k} cos (k t) - (- \frac{y}{k} (e^{y t} \frac{1}{k} sin (k t) - \frac{y}{k} \cdot \int e^{y t} sin (k t) d t))

Therefore

\int e^{y t} sin (k t) d t = - e^{y t} \frac{1}{k} cos (k t) - (- \frac{y}{k} (e^{y t} \frac{1}{k} sin (k t) - \frac{y}{k} \cdot \int e^{y t} sin (k t) d t))

Isolate

\int e^{y t} sin (k t) d t

= - \frac{k e ^{y t} cos ( k t )}{k ^{2} + y ^{2}} + \frac{y e ^{y t} sin ( k t )}{k ^{2} + y ^{2}}

Add a constant to the solution

= - \frac{k e ^{y t} cos ( k t )}{k ^{2} + y ^{2}} + \frac{y e ^{y t} sin ( k t )}{k ^{2} + y ^{2}} + C

What is the integral of e^{yt}*sin(kt) ?
The integral of e^{yt}*sin(kt) is -(ke^{yt}cos(kt))/(k^2+y^2)+(ye^{yt}sin(kt))/(k^2+y^2)+C