What is the integral of sin(t)e^t ?

The integral of sin(t)e^t is -(e^tcos(t))/2+(e^tsin(t))/2+C

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integral of sin(t)e^t

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

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

Solution steps

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

Apply Integration By Parts

: - e^{t} cos (t) - \int - e^{t} cos (t) d t

= - e^{t} cos (t) - \int - e^{t} cos (t) d t

Take the constant out:

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

= - e^{t} cos (t) - (- \int e^{t} cos (t) d t)

Apply Integration By Parts

: e^{t} sin (t) - \int e^{t} sin (t) d t

= - e^{t} cos (t) - (- (e^{t} sin (t) - \int e^{t} sin (t) d t))

Therefore

\int e^{t} sin (t) d t = - e^{t} cos (t) - (- (e^{t} sin (t) - \int e^{t} sin (t) d t))

Isolate

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

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

Add a constant to the solution

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

\frac{12}{3 s - 5}

\frac{d ^{2}}{d x ^{2}} (\frac{6 π ^{2}}{6 - x})

n = 5 \sum \infty \frac{8 ^{n}}{9 ^{n}}

\int \frac{3}{5} - \frac{4}{x} d x

n = 1 \sum \infty n (sin (\frac{1}{n}))

What is the integral of sin(t)e^t ?
The integral of sin(t)e^t is -(e^tcos(t))/2+(e^tsin(t))/2+C