What is the integral of-e^{-2csc(2x)}cot(2x)csc(2x) ?

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

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integral of-e^{-2csc(2x)}cot(2x)csc(2x)

\int - e^{- 2 c s c (2 x)} cot (2 x) csc (2 x) d x

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

Solution steps

\int - e^{- 2 c s c (2 x)} cot (2 x) csc (2 x) d x

Take the constant out:

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

= - \int e^{- 2 c s c (2 x)} cot (2 x) csc (2 x) d x

Apply u-substitution

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

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

Take the constant out:

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

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

Use the common integral:

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

= - \frac{1}{4} e^{u}

Substitute back

u = - 2 csc (2 x)

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

Add a constant to the solution

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

\int \frac{1}{( x - x ^{2} )} d x

y^{^{'}} + \frac{x}{x ^{2} + 1} y = - 4 x

y = (\frac{( x + 2 )}{( x - 3 )})^{2}, at x = 4

\frac{d}{d x} (a (e^{x} + e^{x} x))

h \to 0 lim (h sin (\frac{1}{h}))

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