What is the integral of-5e^{-4sin(2x)}cos(2x) ?

The integral of-5e^{-4sin(2x)}cos(2x) is 5/8 e^{-4sin(2x)}+C

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integral of-5e^{-4sin(2x)}cos(2x)

\int - 5 e^{- 4 s i n (2 x)} cos (2 x) d x

\frac{5}{8} e^{- 4 s i n (2 x)} + C

Solution steps

\int - 5 e^{- 4 s i n (2 x)} cos (2 x) d x

Take the constant out:

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

= - 5 \cdot \int e^{- 4 s i n (2 x)} cos (2 x) d x

Apply u-substitution

: \int e^{- 4 s i n (u)} cos (u) \frac{1}{2} d u

= - 5 \cdot \int e^{- 4 s i n (u)} cos (u) \frac{1}{2} d u

Take the constant out:

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

= - 5 \cdot \frac{1}{2} \cdot \int e^{- 4 s i n (u)} cos (u) d u

Apply u-substitution

: \int - \frac{e ^{v}}{4} d v

= - 5 \cdot \frac{1}{2} \cdot \int - \frac{e ^{v}}{4} d v

Take the constant out:

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

= - 5 \cdot \frac{1}{2} (- \frac{1}{4} \cdot \int e^{v} d v)

Use the common integral:

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

= - 5 \cdot \frac{1}{2} (- \frac{1}{4} e^{v})

Substitute back

= - 5 \cdot \frac{1}{2} (- \frac{1}{4} e^{- 4 s i n (2 x)})

Simplify

- 5 \cdot \frac{1}{2} (- \frac{1}{4} e^{- 4 s i n (2 x)}) : \frac{5}{8} e^{- 4 s i n (2 x)}

= \frac{5}{8} e^{- 4 s i n (2 x)}

Add a constant to the solution

= \frac{5}{8} e^{- 4 s i n (2 x)} + C

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What is the integral of-5e^{-4sin(2x)}cos(2x) ?
The integral of-5e^{-4sin(2x)}cos(2x) is 5/8 e^{-4sin(2x)}+C