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

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

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

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

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

Solution steps

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

Rewrite using trig identities

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

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 (x)} cos (x) sin (x) d x

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 (x)

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

Simplify

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

Add a constant to the solution

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

\int arccosh (x) d x

8 y^{^{'}} + y = 6 t^{2}

\int 1 + 3 x d x

y = cos (x^{2}) 2 x

n = 1 \sum \infty \frac{( e ^{\frac{8}{n}} )}{n}

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