What is the integral of e^{4x}cos(9x) ?

The integral of e^{4x}cos(9x) is (9e^{4x}sin(9x))/(97)+(4e^{4x}cos(9x))/(97)+C

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integral of e^{4x}cos(9x)

\int e^{4 x} cos (9 x) d x

\frac{9 e ^{4 x} sin ( 9 x )}{97} + \frac{4 e ^{4 x} cos ( 9 x )}{97} + C

Solution steps

\int e^{4 x} cos (9 x) d x

Apply Integration By Parts

: \frac{1}{9} e^{4 x} sin (9 x) - \int \frac{4}{9} e^{4 x} sin (9 x) d x

= \frac{1}{9} e^{4 x} sin (9 x) - \int \frac{4}{9} e^{4 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^{4 x} sin (9 x) - \frac{4}{9} \cdot \int e^{4 x} sin (9 x) d x

Apply Integration By Parts

: - \frac{1}{9} e^{4 x} cos (9 x) - \int - \frac{4}{9} e^{4 x} cos (9 x) d x

= \frac{1}{9} e^{4 x} sin (9 x) - \frac{4}{9} (- \frac{1}{9} e^{4 x} cos (9 x) - \int - \frac{4}{9} e^{4 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^{4 x} sin (9 x) - \frac{4}{9} (- \frac{1}{9} e^{4 x} cos (9 x) - (- \frac{4}{9} \cdot \int e^{4 x} cos (9 x) d x))

Therefore

\int e^{4 x} cos (9 x) d x = \frac{1}{9} e^{4 x} sin (9 x) - \frac{4}{9} (- \frac{1}{9} e^{4 x} cos (9 x) - (- \frac{4}{9} \cdot \int e^{4 x} cos (9 x) d x))

Isolate

\int e^{4 x} cos (9 x) d x

= \frac{9 e ^{4 x} sin ( 9 x )}{97} + \frac{4 e ^{4 x} cos ( 9 x )}{97}

Add a constant to the solution

= \frac{9 e ^{4 x} sin ( 9 x )}{97} + \frac{4 e ^{4 x} cos ( 9 x )}{97} + C

\int_{- 2 π}^{2 π} (sin^{2} (x) + 1) d x

\int lo g_{e} (5 x) d x

f (x) = - 2 sin (x)

300 - 3 p

\frac{\partial}{\partial x} (arcsin (y))

What is the integral of e^{4x}cos(9x) ?
The integral of e^{4x}cos(9x) is (9e^{4x}sin(9x))/(97)+(4e^{4x}cos(9x))/(97)+C