What is the integral of e^{2sin(3x)}cos(3x) ?

The integral of e^{2sin(3x)}cos(3x) is 1/6 e^{2sin(3x)}+C

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integral of e^{2sin(3x)}cos(3x)

\int e^{2 s i n (3 x)} cos (3 x) d x

\frac{1}{6} e^{2 s i n (3 x)} + C

Solution steps

\int e^{2 s i n (3 x)} cos (3 x) d x

Apply u-substitution

: \int e^{2 s i n (u)} cos (u) \frac{1}{3} d u

= \int e^{2 s i n (u)} cos (u) \frac{1}{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 e^{2 s i n (u)} cos (u) d u

Apply u-substitution

: \int \frac{e ^{v}}{2} d v

= \frac{1}{3} \cdot \int \frac{e ^{v}}{2} d v

Take the constant out:

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

= \frac{1}{3} \cdot \frac{1}{2} \cdot \int e^{v} d v

Use the common integral:

\int e^{v} d v = e^{v}

= \frac{1}{3} \cdot \frac{1}{2} e^{v}

Substitute back

= \frac{1}{3} \cdot \frac{1}{2} e^{2 s i n (3 x)}

Simplify

\frac{1}{3} \cdot \frac{1}{2} e^{2 s i n (3 x)} : \frac{1}{6} e^{2 s i n (3 x)}

= \frac{1}{6} e^{2 s i n (3 x)}

Add a constant to the solution

= \frac{1}{6} e^{2 s i n (3 x)} + C

\frac{\partial}{\partial x} (2 y^{\frac{1}{2}} x^{\frac{1}{2}})

f (x) = \frac{10}{x}

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\frac{d}{d x} (cos (x) d x)

\int - 4 sin (2 t - \frac{π}{2}) d t

What is the integral of e^{2sin(3x)}cos(3x) ?
The integral of e^{2sin(3x)}cos(3x) is 1/6 e^{2sin(3x)}+C