What is the integral of-6e^{2sin(x)}cos(x) ?

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

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integral of-6e^{2sin(x)}cos(x)

\int - 6 e^{2 s i n (x)} cos (x) d x

- 3 e^{2 s i n (x)} + C

Solution steps

\int - 6 e^{2 s i n (x)} cos (x) d x

Take the constant out:

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

= - 6 \cdot \int e^{2 s i n (x)} cos (x) d x

Apply u-substitution

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

= - 6 \cdot \int \frac{e ^{u}}{2} d u

Take the constant out:

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

= - 6 \cdot \frac{1}{2} \cdot \int e^{u} d u

Use the common integral:

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

= - 6 \cdot \frac{1}{2} e^{u}

Substitute back

u = 2 sin (x)

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

Simplify

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

= - 3 e^{2 s i n (x)}

Add a constant to the solution

= - 3 e^{2 s i n (x)} + C

\int \frac{2 ^{x} - 3 ^{x}}{5 ^{x}} d x

\int \frac{y + 3}{y - 2} d y

\int \frac{1 - x}{2 x} d x

\int - \frac{6}{x ^{2} + 1} d x

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