What is the integral of e^{at}sin(bt) ?

The integral of e^{at}sin(bt) is -(be^{at}cos(bt))/(b^2+a^2)+(ae^{at}sin(bt))/(b^2+a^2)+C

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integral of e^{at}sin(bt)

\int e^{a t} sin (b t) d t

- \frac{b e ^{a t} cos ( b t )}{b ^{2} + a ^{2}} + \frac{a e ^{a t} sin ( b t )}{b ^{2} + a ^{2}} + C

Solution steps

\int e^{a t} sin (b t) d t

Apply Integration By Parts

: - e^{a t} \frac{1}{b} cos (b t) - \int - a e^{a t} \frac{1}{b} cos (b t) d t

= - e^{a t} \frac{1}{b} cos (b t) - \int - a e^{a t} \frac{1}{b} cos (b t) d t

Take the constant out:

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

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

Simplify

- a \frac{1}{b} : - \frac{a}{b}

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

Apply Integration By Parts

: e^{a t} \frac{1}{b} sin (b t) - \int e^{a t} a \frac{1}{b} sin (b t) d t

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

Take the constant out:

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

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

Simplify

a \frac{1}{b} : \frac{a}{b}

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

Therefore

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

Isolate

\int e^{a t} sin (b t) d t

= - \frac{b e ^{a t} cos ( b t )}{b ^{2} + a ^{2}} + \frac{a e ^{a t} sin ( b t )}{b ^{2} + a ^{2}}

Add a constant to the solution

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

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1 + sin^{2} (x)

\int 1 + sin (x) cos (x) d x

\int x + \frac{2}{sin ( x )} d x

What is the integral of e^{at}sin(bt) ?
The integral of e^{at}sin(bt) is -(be^{at}cos(bt))/(b^2+a^2)+(ae^{at}sin(bt))/(b^2+a^2)+C