What is the integral of e^{2t}cos(5t) ?

The integral of e^{2t}cos(5t) is (5e^{2t}sin(5t))/(29)+(2e^{2t}cos(5t))/(29)+C

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integral of e^{2t}cos(5t)

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

\frac{5 e ^{2 t} sin ( 5 t )}{29} + \frac{2 e ^{2 t} cos ( 5 t )}{29} + C

Solution steps

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

Apply Integration By Parts

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

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

Take the constant out:

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

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

Apply Integration By Parts

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

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

Take the constant out:

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

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

Therefore

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

Isolate

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

= \frac{5 e ^{2 t} sin ( 5 t )}{29} + \frac{2 e ^{2 t} cos ( 5 t )}{29}

Add a constant to the solution

= \frac{5 e ^{2 t} sin ( 5 t )}{29} + \frac{2 e ^{2 t} cos ( 5 t )}{29} + C

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x \to 0 + lim (x^{2} ln (x))

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What is the integral of e^{2t}cos(5t) ?
The integral of e^{2t}cos(5t) is (5e^{2t}sin(5t))/(29)+(2e^{2t}cos(5t))/(29)+C