What is the integral of e^{1+cos(3x)}sin(3x) ?

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

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integral of e^{1+cos(3x)}sin(3x)

\int e^{1 + c o s (3 x)} sin (3 x) d x

- \frac{1}{3} e^{1 + c o s (3 x)} + C

Solution steps

\int e^{1 + c o s (3 x)} sin (3 x) d x

e^{1 + c o s (3 x)} sin (3 x) = e^{1} e^{c o s (3 x)} sin (3 x)

= \int e^{1} e^{c o s (3 x)} sin (3 x) d x

Take the constant out:

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

= e^{1} \cdot \int e^{c o s (3 x)} sin (3 x) d x

Simplify

= e \cdot \int e^{c o s (3 x)} sin (3 x) d x

Apply u-substitution

: \int e^{c o s (u)} sin (u) \frac{1}{3} d u

= e \cdot \int e^{c o s (u)} sin (u) \frac{1}{3} d u

Take the constant out:

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

= e \frac{1}{3} \cdot \int e^{c o s (u)} sin (u) d u

Apply u-substitution

: \int - e^{v} d v

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

Take the constant out:

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

= e \frac{1}{3} (- \int e^{v} d v)

Use the common integral:

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

= e \frac{1}{3} (- e^{v})

Substitute back

= e \frac{1}{3} (- e^{c o s (3 x)})

Simplify

e \frac{1}{3} (- e^{c o s (3 x)}) : - \frac{1}{3} e^{1 + c o s (3 x)}

= - \frac{1}{3} e^{1 + c o s (3 x)}

Add a constant to the solution

= - \frac{1}{3} e^{1 + c o s (3 x)} + C

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