What is the integral of e^{cos(40t)}sin(40t) ?

The integral of e^{cos(40t)}sin(40t) is -1/40 e^{cos(40t)}+C

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integral of e^{cos(40t)}sin(40t)

\int e^{c o s (40 t)} sin (40 t) d t

- \frac{1}{40} e^{c o s (40 t)} + C

Solution steps

\int e^{c o s (40 t)} sin (40 t) d t

Apply u-substitution

: \int e^{c o s (u)} sin (u) \frac{1}{40} d u

= \int e^{c o s (u)} sin (u) \frac{1}{40} d u

Take the constant out:

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

= \frac{1}{40} \cdot \int e^{c o s (u)} sin (u) d u

Apply u-substitution

: \int - e^{v} d v

= \frac{1}{40} \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}{40} (- \int e^{v} d v)

Use the common integral:

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

= \frac{1}{40} (- e^{v})

Substitute back

= \frac{1}{40} (- e^{c o s (40 t)})

Simplify

= - \frac{1}{40} e^{c o s (40 t)}

Add a constant to the solution

= - \frac{1}{40} e^{c o s (40 t)} + C

\frac{d}{d x} (\frac{6 x}{x ^{2} + 36})

e^{- 3}

t \to 1 + lim (t^{\frac{t}{1 - t}})

\frac{d}{d x} (\frac{sin ( x )}{tan ( x )})

x \to \frac{π}{5} - lim (cot (5 x))

What is the integral of e^{cos(40t)}sin(40t) ?
The integral of e^{cos(40t)}sin(40t) is -1/40 e^{cos(40t)}+C