What is the integral of e^{2x}cos(9x) ?

The integral of e^{2x}cos(9x) is (9e^{2x}sin(9x))/(85)+(2e^{2x}cos(9x))/(85)+C

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integral of e^{2x}cos(9x)

\int e^{2 x} cos (9 x) d x

\frac{9 e ^{2 x} sin ( 9 x )}{85} + \frac{2 e ^{2 x} cos ( 9 x )}{85} + C

Solution steps

\int e^{2 x} cos (9 x) d x

Apply Integration By Parts

: \frac{1}{9} e^{2 x} sin (9 x) - \int \frac{2}{9} e^{2 x} sin (9 x) d x

= \frac{1}{9} e^{2 x} sin (9 x) - \int \frac{2}{9} e^{2 x} sin (9 x) d x

Take the constant out:

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

= \frac{1}{9} e^{2 x} sin (9 x) - \frac{2}{9} \cdot \int e^{2 x} sin (9 x) d x

Apply Integration By Parts

: - \frac{1}{9} e^{2 x} cos (9 x) - \int - \frac{2}{9} e^{2 x} cos (9 x) d x

= \frac{1}{9} e^{2 x} sin (9 x) - \frac{2}{9} (- \frac{1}{9} e^{2 x} cos (9 x) - \int - \frac{2}{9} e^{2 x} cos (9 x) d x)

Take the constant out:

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

= \frac{1}{9} e^{2 x} sin (9 x) - \frac{2}{9} (- \frac{1}{9} e^{2 x} cos (9 x) - (- \frac{2}{9} \cdot \int e^{2 x} cos (9 x) d x))

Therefore

\int e^{2 x} cos (9 x) d x = \frac{1}{9} e^{2 x} sin (9 x) - \frac{2}{9} (- \frac{1}{9} e^{2 x} cos (9 x) - (- \frac{2}{9} \cdot \int e^{2 x} cos (9 x) d x))

Isolate

\int e^{2 x} cos (9 x) d x

= \frac{9 e ^{2 x} sin ( 9 x )}{85} + \frac{2 e ^{2 x} cos ( 9 x )}{85}

Add a constant to the solution

= \frac{9 e ^{2 x} sin ( 9 x )}{85} + \frac{2 e ^{2 x} cos ( 9 x )}{85} + C

\frac{d y}{d x}, y = \frac{2}{x}

\int e^{- 4 x^{2}} d x

n = 1 \sum \infty (cos (0.3))^{n}

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

\frac{d y}{d t} + y = t e^{t}

What is the integral of e^{2x}cos(9x) ?
The integral of e^{2x}cos(9x) is (9e^{2x}sin(9x))/(85)+(2e^{2x}cos(9x))/(85)+C