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

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

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

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

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

Solution steps

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

Apply u-substitution

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

= \int cos (u) e^{2 s i n (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 cos (u) e^{2 s i n (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{d y}{d x} = x x - 30

\frac{d}{d x} (2 sin (x) \cdot cos (x))

0

x \to 0 lim (5 sin (x) ln (x))

\frac{d}{d x} (4 - \frac{35}{8 x ^{\frac{1}{8}}})

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