What is the integral of e^{cos(3t)}sin(3t) ?

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

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integral of e^{cos(3t)}sin(3t)

\int e^{c o s (3 t)} sin (3 t) d t

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

Solution steps

\int e^{c o s (3 t)} sin (3 t) d t

Apply u-substitution

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

= \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

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

Apply u-substitution

: \int - e^{v} d v

= \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

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

Use the common integral:

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

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

Substitute back

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

Simplify

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

Add a constant to the solution

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

y^{^{''}} + 2 y^{^{'}} + 12 y = 0, y (0) = 1, y^{^{'}} (0) = 0

x \to 1 - lim (3 x^{2} - 4 x + 4)

\int \frac{ln ( x + 1 )}{x + 1} d x

y^{^{'}} = y (y - 3), y (0) = a

n = 0 \sum \infty \frac{n x ^{n}}{n + 2}

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