What is the integral of e^{cos(16t)}sin(16t) ?

The integral of e^{cos(16t)}sin(16t) is -1/16 e^{cos(16t)}+C

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integral of e^{cos(16t)}sin(16t)

\int e^{c o s (16 t)} sin (16 t) d t

- \frac{1}{16} e^{c o s (16 t)} + C

Solution steps

\int e^{c o s (16 t)} sin (16 t) d t

Apply u-substitution

: \int e^{c o s (u)} sin (u) \frac{1}{16} d u

= \int e^{c o s (u)} sin (u) \frac{1}{16} d u

Take the constant out:

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

= \frac{1}{16} \cdot \int e^{c o s (u)} sin (u) d u

Apply u-substitution

: \int - e^{v} d v

= \frac{1}{16} \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}{16} (- \int e^{v} d v)

Use the common integral:

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

= \frac{1}{16} (- e^{v})

Substitute back

= \frac{1}{16} (- e^{c o s (16 t)})

Simplify

= - \frac{1}{16} e^{c o s (16 t)}

Add a constant to the solution

= - \frac{1}{16} e^{c o s (16 t)} + C

\int y^{2} cos (x) d x

y = sin (x), y = sin^{3} (x), x = 0, x = \frac{π}{2}

y^{^{''}} + 16 y = - 16 sin (4 t)

\int_{s}^{g} \frac{1}{1 - 2 \cdot \frac{m}{r}} d r

\frac{d}{d x} (x^{2} - x - 10)

What is the integral of e^{cos(16t)}sin(16t) ?
The integral of e^{cos(16t)}sin(16t) is -1/16 e^{cos(16t)}+C