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

The integral of e^{cos(2x)}sin(x)cos(x) is -1/4 e^{cos(2x)}+C

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

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

- \frac{1}{4} e^{c o s (2 x)} + C

Solution steps

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

Rewrite using trig identities

= \int \frac{1}{2} e^{c o s (2 x)} sin (2 x) d x

Take the constant out:

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

= \frac{1}{2} \cdot \int e^{c o s (2 x)} sin (2 x) d x

Apply u-substitution

: \int e^{c o s (u)} sin (u) \frac{1}{2} d u

= \frac{1}{2} \cdot \int e^{c o s (u)} sin (u) \frac{1}{2} d u

Take the constant out:

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

= \frac{1}{2} \cdot \frac{1}{2} \cdot \int e^{c o s (u)} sin (u) d u

Apply u-substitution

: \int - e^{v} d v

= \frac{1}{2} \cdot \frac{1}{2} \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}{2} \cdot \frac{1}{2} (- \int e^{v} d v)

Use the common integral:

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

= \frac{1}{2} \cdot \frac{1}{2} (- e^{v})

Substitute back

= \frac{1}{2} \cdot \frac{1}{2} (- e^{c o s (2 x)})

Simplify

\frac{1}{2} \cdot \frac{1}{2} (- e^{c o s (2 x)}) : - \frac{1}{4} e^{c o s (2 x)}

= - \frac{1}{4} e^{c o s (2 x)}

Add a constant to the solution

= - \frac{1}{4} e^{c o s (2 x)} + C

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

(t^{2} + x^{2}) d t + (2 t x - x) d x = 0

\frac{d}{d x} (x^{4} - 4 x^{2} - 10 x)

x \to 0 - lim (- 7 x^{2} + 5 x)

y = \frac{6 x - 1}{8 x + 3}

What is the integral of e^{cos(2x)}sin(x)cos(x) ?
The integral of e^{cos(2x)}sin(x)cos(x) is -1/4 e^{cos(2x)}+C