What is the integral of 5e^tsin(2t) ?

The integral of 5e^tsin(2t) is -2e^tcos(2t)+e^tsin(2t)+C

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integral of 5e^tsin(2t)

\int 5 e^{t} sin (2 t) d t

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

Solution steps

\int 5 e^{t} sin (2 t) d t

Take the constant out:

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

= 5 \cdot \int e^{t} sin (2 t) d t

Apply Integration By Parts

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

= 5 (- \frac{1}{2} e^{t} cos (2 t) - \int - \frac{1}{2} e^{t} cos (2 t) d t)

Take the constant out:

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

= 5 (- \frac{1}{2} e^{t} cos (2 t) - (- \frac{1}{2} \cdot \int e^{t} cos (2 t) d t))

Apply Integration By Parts

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

= 5 (- \frac{1}{2} e^{t} cos (2 t) - (- \frac{1}{2} (e^{t} \frac{1}{2} sin (2 t) - \int e^{t} \frac{1}{2} sin (2 t) d t)))

Take the constant out:

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

= 5 (- \frac{1}{2} e^{t} cos (2 t) - (- \frac{1}{2} (e^{t} \frac{1}{2} sin (2 t) - \frac{1}{2} \cdot \int e^{t} sin (2 t) d t)))

Therefore

5 \cdot \int e^{t} sin (2 t) d t = 5 (- \frac{1}{2} e^{t} cos (2 t) - (- \frac{1}{2} (e^{t} \frac{1}{2} sin (2 t) - \frac{1}{2} \cdot \int e^{t} sin (2 t) d t)))

Isolate

\int e^{t} sin (2 t) d t

= 5 (- \frac{2 e ^{t} cos ( 2 t )}{5} + \frac{e ^{t} sin ( 2 t )}{5})

Simplify

5 (- \frac{2 e ^{t} cos ( 2 t )}{5} + \frac{e ^{t} sin ( 2 t )}{5}) : - 2 e^{t} cos (2 t) + e^{t} sin (2 t)

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

Add a constant to the solution

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

x \to 8 lim (lo g_{2} (x))

\frac{d}{d x} (\frac{ln ( 8 ^{x} + 5 )}{ln ( 8 ^{x + 1} + 5 )})

x \to - 3 lim (x^{2} + 3 x)

\int ln (4 x^{2}) d x

x \to \frac{3 π}{2} + lim (tan (x))

What is the integral of 5e^tsin(2t) ?
The integral of 5e^tsin(2t) is -2e^tcos(2t)+e^tsin(2t)+C