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

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

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

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

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

Solution steps

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

Apply Integration By Parts

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

Take the constant out:

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

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

Apply Integration By Parts

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

Take the constant out:

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

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

Therefore

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

Isolate

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

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

Add a constant to the solution

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

2 t^{2} (\frac{d x}{d t}) = 3 t x + x^{2}

f (x) = 5 sin (x)

d er i v a t i v e f (x) = e^{- 5 x^{4}}

a re a y = 4 x + 2, 0, 4

in v erse l a pl a ce \frac{1}{s + 1}

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