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

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

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

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

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

Solution steps

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

Apply u-substitution

: \int - e^{- 2 u} u d u

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

Take the constant out:

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

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

Apply u-substitution

: \int \frac{e ^{v} v}{4} d v

= - \int \frac{e ^{v} v}{4} d v

Take the constant out:

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

= - \frac{1}{4} \cdot \int e^{v} v d v

Apply Integration By Parts

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

= - \frac{1}{4} (e^{v} v - \int e^{v} d v)

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

= - \frac{1}{4} (e^{v} v - e^{v})

Substitute back

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

Simplify

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

Add a constant to the solution

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

y^{^{'}} = 2 x (x - 4)

x \to π + lim (csc (x))

\int tan^{2} (t) d t

\int \frac{x ^{3} + 93 x + 70}{x ^{3} + 100 x} d x

y = x^{3} - \frac{x ^{2}}{16} + 4 x + 5

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