What is the integral of 8e^{2x}sin(2x) ?

The integral of 8e^{2x}sin(2x) is 8((e^{2x}sin(2x))/4-(e^{2x}cos(2x))/4)+C

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integral of 8e^{2x}sin(2x)

\int 8 e^{2 x} sin (2 x) d x

8 (\frac{e ^{2 x} sin ( 2 x )}{4} - \frac{e ^{2 x} cos ( 2 x )}{4}) + C

Solution steps

\int 8 e^{2 x} sin (2 x) d x

Take the constant out:

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

= 8 \cdot \int e^{2 x} sin (2 x) d x

Apply Integration By Parts

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

= 8 (- \frac{1}{2} e^{2 x} cos (2 x) - \int - e^{2 x} cos (2 x) d x)

Take the constant out:

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

= 8 (- \frac{1}{2} e^{2 x} cos (2 x) - (- \int e^{2 x} cos (2 x) d x))

Apply Integration By Parts

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

= 8 (- \frac{1}{2} e^{2 x} cos (2 x) - (- (\frac{1}{2} e^{2 x} sin (2 x) - \int e^{2 x} sin (2 x) d x)))

Therefore

8 \cdot \int e^{2 x} sin (2 x) d x = 8 (- \frac{1}{2} e^{2 x} cos (2 x) - (- (\frac{1}{2} e^{2 x} sin (2 x) - \int e^{2 x} sin (2 x) d x)))

Isolate

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

= 8 (\frac{e ^{2 x} sin ( 2 x )}{4} - \frac{e ^{2 x} cos ( 2 x )}{4})

Add a constant to the solution

= 8 (\frac{e ^{2 x} sin ( 2 x )}{4} - \frac{e ^{2 x} cos ( 2 x )}{4}) + C

\frac{d}{d x} (- 4 cot (x) csc (x))

\int - \frac{sin ( 8 x )}{8} d x

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7 + 6 sin (x^{2})

What is the integral of 8e^{2x}sin(2x) ?
The integral of 8e^{2x}sin(2x) is 8((e^{2x}sin(2x))/4-(e^{2x}cos(2x))/4)+C