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

The integral of cos(2x)e^{esin(2x)} is 1/2 e^{esin(2x)-1}+C

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integral of cos(2x)e^{esin(2x)}

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

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

Solution steps

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

Apply u-substitution

: \int \frac{e ^{u - 1}}{2} d u

= \int \frac{e ^{u - 1}}{2} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int e^{u - 1} d u

e^{u - 1} = e^{- 1} e^{u}

= \frac{1}{2} \cdot \int e^{- 1} e^{u} d u

Take the constant out:

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

= \frac{1}{2} e^{- 1} \cdot \int e^{u} d u

Apply exponent rule:

a^{- 1} = \frac{1}{a}

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

Use the common integral:

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

= \frac{1}{2} \cdot \frac{1}{e} e^{u}

Substitute back

u = e sin (2 x)

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

Simplify

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

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

Add a constant to the solution

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

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