What is the integral of e^{3x}(sin(3x)+cos(3x)) ?

The integral of e^{3x}(sin(3x)+cos(3x)) is 1/3 e^{3x}sin(3x)+C

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integral of e^{3x}(sin(3x)+cos(3x))

\int e^{3 x} (sin (3 x) + cos (3 x)) d x

\frac{1}{3} e^{3 x} sin (3 x) + C

Solution steps

\int e^{3 x} (sin (3 x) + cos (3 x)) d x

Expand

e^{3 x} (sin (3 x) + cos (3 x)) : e^{3 x} sin (3 x) + e^{3 x} cos (3 x)

= \int e^{3 x} sin (3 x) + e^{3 x} cos (3 x) d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int e^{3 x} sin (3 x) d x + \int e^{3 x} cos (3 x) d x

\int e^{3 x} sin (3 x) d x = - \frac{e ^{3 x} cos ( 3 x )}{6} + \frac{e ^{3 x} sin ( 3 x )}{6}

\int e^{3 x} cos (3 x) d x = \frac{e ^{3 x} sin ( 3 x )}{6} + \frac{e ^{3 x} cos ( 3 x )}{6}

= (- \frac{e ^{3 x} cos ( 3 x )}{6} + \frac{e ^{3 x} sin ( 3 x )}{6}) + \frac{e ^{3 x} sin ( 3 x )}{6} + \frac{e ^{3 x} cos ( 3 x )}{6}

Simplify

(- \frac{e ^{3 x} cos ( 3 x )}{6} + \frac{e ^{3 x} sin ( 3 x )}{6}) + \frac{e ^{3 x} sin ( 3 x )}{6} + \frac{e ^{3 x} cos ( 3 x )}{6} : \frac{1}{3} e^{3 x} sin (3 x)

= \frac{1}{3} e^{3 x} sin (3 x)

Add a constant to the solution

= \frac{1}{3} e^{3 x} sin (3 x) + C

\int (x^{2} - \frac{3}{x}) d x

n = 0 \sum \infty \frac{1}{3 n}

y = \frac{( 3 x - 6 )}{( x ^{2} + 4 )}

x^{2} + y^{3} = 12

\frac{d}{d x} (9 e^{x} + 5)

What is the integral of e^{3x}(sin(3x)+cos(3x)) ?
The integral of e^{3x}(sin(3x)+cos(3x)) is 1/3 e^{3x}sin(3x)+C