What is the integral of e^{9x}cos(5x) ?

The integral of e^{9x}cos(5x) is (5e^{9x}sin(5x))/(106)+(9e^{9x}cos(5x))/(106)+C

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integral of e^{9x}cos(5x)

\int e^{9 x} cos (5 x) d x

\frac{5 e ^{9 x} sin ( 5 x )}{106} + \frac{9 e ^{9 x} cos ( 5 x )}{106} + C

Solution steps

\int e^{9 x} cos (5 x) d x

Apply Integration By Parts

= \frac{1}{5} e^{9 x} sin (5 x) - \int \frac{9}{5} e^{9 x} sin (5 x) d x

Take the constant out:

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

= \frac{1}{5} e^{9 x} sin (5 x) - \frac{9}{5} \cdot \int e^{9 x} sin (5 x) d x

Apply Integration By Parts

= \frac{1}{5} e^{9 x} sin (5 x) - \frac{9}{5} (- \frac{1}{5} e^{9 x} cos (5 x) - \int - \frac{9}{5} e^{9 x} cos (5 x) d x)

Take the constant out:

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

= \frac{1}{5} e^{9 x} sin (5 x) - \frac{9}{5} (- \frac{1}{5} e^{9 x} cos (5 x) - (- \frac{9}{5} \cdot \int e^{9 x} cos (5 x) d x))

Therefore

\int e^{9 x} cos (5 x) d x = \frac{1}{5} e^{9 x} sin (5 x) - \frac{9}{5} (- \frac{1}{5} e^{9 x} cos (5 x) - (- \frac{9}{5} \cdot \int e^{9 x} cos (5 x) d x))

Isolate

\int e^{9 x} cos (5 x) d x

= \frac{5 e ^{9 x} sin ( 5 x )}{106} + \frac{9 e ^{9 x} cos ( 5 x )}{106}

Add a constant to the solution

= \frac{5 e ^{9 x} sin ( 5 x )}{106} + \frac{9 e ^{9 x} cos ( 5 x )}{106} + C

d er i v a t i v e y = - x^{3} + 2 x

ma c l a u r in sin (\frac{π}{6} + x)

\int 1^{\frac{1}{2}}

\frac{\partial}{\partial x} (x^{7} y z^{2} + 7 yz)

a re a x = y, x = - y^{2} + 6 y

What is the integral of e^{9x}cos(5x) ?
The integral of e^{9x}cos(5x) is (5e^{9x}sin(5x))/(106)+(9e^{9x}cos(5x))/(106)+C