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

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

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

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

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

Solution steps

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

Apply u-substitution

: \int \frac{sin ( u )}{2} d u

= 2 \cdot \int \frac{sin ( u )}{2} d u

Take the constant out:

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

= 2 \cdot \frac{1}{2} \cdot \int sin (u) d u

Use the common integral:

\int sin (u) d u = - cos (u)

= 2 \cdot \frac{1}{2} (- cos (u))

Substitute back

u = e^{2 x}

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

Simplify

2 \cdot \frac{1}{2} (- cos (e^{2 x})) : - cos (e^{2 x})

= - cos (e^{2 x})

Add a constant to the solution

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

x y^{^{'}} + y = - 6 x y^{2}

\frac{3}{2} x^{\frac{2}{3}}

\frac{\partial}{\partial y} (a ln (x^{2} + y^{2}))

x \to 3 - lim (\frac{1}{6 - 2 x})

\frac{1}{s ^{2} - 2 s + 5}

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