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

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

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

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

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

Solution steps

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

Apply u-substitution

: \int \frac{u sin ( u )}{3} d u

= \int \frac{u sin ( u )}{3} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int u sin (u) d u

Apply Integration By Parts

: - u cos (u) - \int - cos (u) d u

= \frac{1}{3} (- u cos (u) - \int - cos (u) d u)

\int - cos (u) d u = - sin (u)

= \frac{1}{3} (- u cos (u) - (- sin (u)))

Substitute back

u = e^{3 x}

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

Simplify

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

Add a constant to the solution

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

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

\int e^{x} 25 + e^{x} d x

\frac{d y}{d x} = \frac{x ^{2} + y ^{2}}{2 x y}

\frac{d}{d x} (cot (x) - sin (2 x))

y = x, x^{2}, [0, 1]

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