What is the integral of e^{cos(t)}sin(2t) ?

The integral of e^{cos(t)}sin(2t) is -2(e^{cos(t)}cos(t)-e^{cos(t)})+C

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integral of e^{cos(t)}sin(2t)

\int e^{c o s (t)} sin (2 t) d t

- 2 (e^{c o s (t)} cos (t) - e^{c o s (t)}) + C

Solution steps

\int e^{c o s (t)} sin (2 t) d t

Rewrite using trig identities

= \int e^{c o s (t)} \cdot 2 cos (t) sin (t) d t

Take the constant out:

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

= 2 \cdot \int e^{c o s (t)} cos (t) sin (t) d t

Apply u-substitution

: \int - e^{u} u d u

= 2 \cdot \int - e^{u} u d u

Take the constant out:

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

= 2 (- \int e^{u} u d u)

Apply Integration By Parts

: e^{u} u - \int e^{u} d u

= 2 (- (e^{u} u - \int e^{u} d u))

Use the common integral:

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

= 2 (- (e^{u} u - e^{u}))

Substitute back

u = cos (t)

= 2 (- (e^{c o s (t)} cos (t) - e^{c o s (t)}))

Simplify

= - 2 (e^{c o s (t)} cos (t) - e^{c o s (t)})

Add a constant to the solution

= - 2 (e^{c o s (t)} cos (t) - e^{c o s (t)}) + C

\frac{6}{9 - x}

\int cos (v) d v

\int 7 sin^{3} (x) cos (x) d x

\frac{d}{d x} (\frac{1}{6} ln (e^{2 x} - e^{x} + 1))

x h (x)

What is the integral of e^{cos(t)}sin(2t) ?
The integral of e^{cos(t)}sin(2t) is -2(e^{cos(t)}cos(t)-e^{cos(t)})+C