What is the integral of e^{cos(4t)}sin(4t) ?

The integral of e^{cos(4t)}sin(4t) is -1/4 e^{cos(4t)}+C

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integral of e^{cos(4t)}sin(4t)

\int e^{c o s (4 t)} sin (4 t) d t

- \frac{1}{4} e^{c o s (4 t)} + C

Solution steps

\int e^{c o s (4 t)} sin (4 t) d t

Apply u-substitution

: \int e^{c o s (u)} sin (u) \frac{1}{4} d u

= \int e^{c o s (u)} sin (u) \frac{1}{4} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int e^{c o s (u)} sin (u) d u

Apply u-substitution

: \int - e^{v} d v

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

Take the constant out:

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

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

Use the common integral:

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

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

Substitute back

= \frac{1}{4} (- e^{c o s (4 t)})

Simplify

= - \frac{1}{4} e^{c o s (4 t)}

Add a constant to the solution

= - \frac{1}{4} e^{c o s (4 t)} + C

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What is the integral of e^{cos(4t)}sin(4t) ?
The integral of e^{cos(4t)}sin(4t) is -1/4 e^{cos(4t)}+C