What is the integral of e^x5sin(2x) ?

The integral of e^x5sin(2x) is -2e^xcos(2x)+e^xsin(2x)+C

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integral of e^x5sin(2x)

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

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

Solution steps

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

Take the constant out:

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

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

Apply Integration By Parts

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

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

Take the constant out:

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

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

Apply Integration By Parts

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

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

Take the constant out:

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

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

Therefore

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

Isolate

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

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

Simplify

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

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

Add a constant to the solution

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

x \to 9 lim (\frac{x ^{2} - 17 x + 72}{x - 9})

(t - 1) (t - 1)

9 e^{x} cos (x)

\frac{s + 3}{( s - 2 ) ( s + 1 )}

x \to 1 lim (2 - x^{2})

What is the integral of e^x5sin(2x) ?
The integral of e^x5sin(2x) is -2e^xcos(2x)+e^xsin(2x)+C