What is the integral of e^{6x}cos(e^{3x}) ?

The integral of e^{6x}cos(e^{3x}) is 1/3 e^{3x}sin(e^{3x})+1/3 cos(e^{3x})+C

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integral of e^{6x}cos(e^{3x})

\int e^{6 x} cos (e^{3 x}) d x

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

Solution steps

\int e^{6 x} cos (e^{3 x}) d x

Apply Integration By Parts

: \frac{1}{6} e^{6 x} cos (e^{3 x}) - \int - \frac{1}{2} e^{9 x} sin (e^{3 x}) d x

= \frac{1}{6} e^{6 x} cos (e^{3 x}) - \int - \frac{1}{2} e^{9 x} sin (e^{3 x}) d x

\int - \frac{1}{2} e^{9 x} sin (e^{3 x}) d x = - \frac{1}{6} (- e^{6 x} cos (e^{3 x}) + 2 (e^{3 x} sin (e^{3 x}) + cos (e^{3 x})))

= \frac{1}{6} e^{6 x} cos (e^{3 x}) - (- \frac{1}{6} (- e^{6 x} cos (e^{3 x}) + 2 (e^{3 x} sin (e^{3 x}) + cos (e^{3 x}))))

Simplify

\frac{1}{6} e^{6 x} cos (e^{3 x}) - (- \frac{1}{6} (- e^{6 x} cos (e^{3 x}) + 2 (e^{3 x} sin (e^{3 x}) + cos (e^{3 x})))) : \frac{1}{3} e^{3 x} sin (e^{3 x}) + \frac{1}{3} cos (e^{3 x})

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

Add a constant to the solution

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

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What is the integral of e^{6x}cos(e^{3x}) ?
The integral of e^{6x}cos(e^{3x}) is 1/3 e^{3x}sin(e^{3x})+1/3 cos(e^{3x})+C