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

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

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

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

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

Solution steps

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

Apply u-substitution

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

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

Use the common integral:

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

= - \frac{1}{2} e^{u}

Substitute back

u = csc (2 x)

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

Add a constant to the solution

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

16 y^{^{''}} - 8 y^{^{'}} + y = 0

4 e^{- t} + 2

\int \frac{1}{1 + v ^{2}} d v

\int \frac{( ln ( x ) ) ^{5}}{x} d x

\frac{d}{d x} (5 sec (5 x) tan (5 x))

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