What is the integral of e^{20t}6sin(2t) ?

The integral of e^{20t}6sin(2t) is 6(-(e^{20t}cos(2t))/(202)+(5e^{20t}sin(2t))/(101))+C

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integral of e^{20t}6sin(2t)

\int e^{20 t} 6 sin (2 t) d t

6 (- \frac{e ^{20 t} cos ( 2 t )}{202} + \frac{5 e ^{20 t} sin ( 2 t )}{101}) + C

Solution steps

\int e^{20 t} \cdot 6 sin (2 t) d t

Take the constant out:

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

= 6 \cdot \int e^{20 t} sin (2 t) d t

Apply Integration By Parts

: - \frac{1}{2} e^{20 t} cos (2 t) - \int - 10 e^{20 t} cos (2 t) d t

= 6 (- \frac{1}{2} e^{20 t} cos (2 t) - \int - 10 e^{20 t} cos (2 t) d t)

Take the constant out:

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

= 6 (- \frac{1}{2} e^{20 t} cos (2 t) - (- 10 \cdot \int e^{20 t} cos (2 t) d t))

Apply Integration By Parts

: \frac{1}{2} e^{20 t} sin (2 t) - \int 10 e^{20 t} sin (2 t) d t

= 6 (- \frac{1}{2} e^{20 t} cos (2 t) - (- 10 (\frac{1}{2} e^{20 t} sin (2 t) - \int 10 e^{20 t} sin (2 t) d t)))

Take the constant out:

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

= 6 (- \frac{1}{2} e^{20 t} cos (2 t) - (- 10 (\frac{1}{2} e^{20 t} sin (2 t) - 10 \cdot \int e^{20 t} sin (2 t) d t)))

Therefore

6 \cdot \int e^{20 t} sin (2 t) d t = 6 (- \frac{1}{2} e^{20 t} cos (2 t) - (- 10 (\frac{1}{2} e^{20 t} sin (2 t) - 10 \cdot \int e^{20 t} sin (2 t) d t)))

Isolate

\int e^{20 t} sin (2 t) d t

= 6 (- \frac{e ^{20 t} cos ( 2 t )}{202} + \frac{5 e ^{20 t} sin ( 2 t )}{101})

Add a constant to the solution

= 6 (- \frac{e ^{20 t} cos ( 2 t )}{202} + \frac{5 e ^{20 t} sin ( 2 t )}{101}) + C

\frac{d x}{d t} + 0.1 x = 10

\int \frac{6 s ^{2} - 2}{s ^{3}} d s

\frac{1}{2 x + 5}

2 x^{2} - 3 x + 5

f (x) = \frac{sin ( 5 x )}{3 x}

What is the integral of e^{20t}6sin(2t) ?
The integral of e^{20t}6sin(2t) is 6(-(e^{20t}cos(2t))/(202)+(5e^{20t}sin(2t))/(101))+C