What is the integral of 4sin(wt)e^t ?

The integral of 4sin(wt)e^t is 4(-(we^tcos(wt))/(w^2+1)+(e^tsin(wt))/(w^2+1))+C

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integral of 4sin(wt)e^t

\int 4 sin (wt) e^{t} d t

4 (- \frac{w e ^{t} cos ( wt )}{w ^{2} + 1} + \frac{e ^{t} sin ( wt )}{w ^{2} + 1}) + C

Solution steps

\int 4 sin (wt) e^{t} d t

Take the constant out:

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

= 4 \cdot \int sin (wt) e^{t} d t

Apply Integration By Parts

: - e^{t} \frac{1}{w} cos (wt) - \int - e^{t} \frac{1}{w} cos (wt) d t

= 4 (- e^{t} \frac{1}{w} cos (wt) - \int - e^{t} \frac{1}{w} cos (wt) d t)

Take the constant out:

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

= 4 (- e^{t} \frac{1}{w} cos (wt) - (- \frac{1}{w} \cdot \int e^{t} cos (wt) d t))

Apply Integration By Parts

: e^{t} \frac{1}{w} sin (wt) - \int e^{t} \frac{1}{w} sin (wt) d t

= 4 (- e^{t} \frac{1}{w} cos (wt) - (- \frac{1}{w} (e^{t} \frac{1}{w} sin (wt) - \int e^{t} \frac{1}{w} sin (wt) d t)))

Take the constant out:

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

= 4 (- e^{t} \frac{1}{w} cos (wt) - (- \frac{1}{w} (e^{t} \frac{1}{w} sin (wt) - \frac{1}{w} \cdot \int e^{t} sin (wt) d t)))

Therefore

4 \cdot \int e^{t} sin (wt) d t = 4 (- e^{t} \frac{1}{w} cos (wt) - (- \frac{1}{w} (e^{t} \frac{1}{w} sin (wt) - \frac{1}{w} \cdot \int e^{t} sin (wt) d t)))

Isolate

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

= 4 (- \frac{w e ^{t} cos ( wt )}{w ^{2} + 1} + \frac{e ^{t} sin ( wt )}{w ^{2} + 1})

Add a constant to the solution

= 4 (- \frac{w e ^{t} cos ( wt )}{w ^{2} + 1} + \frac{e ^{t} sin ( wt )}{w ^{2} + 1}) + C

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\int \frac{1}{4} d x

\frac{d}{d x} ((\frac{3}{4})^{- \frac{4}{3}})

\frac{\partial}{\partial x} (2 e^{x yz} x z)

x^{2} y^{^{'}} = 3 x y + 4 x^{6}

What is the integral of 4sin(wt)e^t ?
The integral of 4sin(wt)e^t is 4(-(we^tcos(wt))/(w^2+1)+(e^tsin(wt))/(w^2+1))+C