What is the integral of 4e^{4x}sin(e^{4x}) ?

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

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integral of 4e^{4x}sin(e^{4x})

\int 4 e^{4 x} sin (e^{4 x}) d x

- cos (e^{4 x}) + C

Solution steps

\int 4 e^{4 x} sin (e^{4 x}) d x

Take the constant out:

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

= 4 \cdot \int e^{4 x} sin (e^{4 x}) d x

Apply u-substitution

: \int \frac{sin ( u )}{4} d u

= 4 \cdot \int \frac{sin ( u )}{4} d u

Take the constant out:

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

= 4 \cdot \frac{1}{4} \cdot \int sin (u) d u

Use the common integral:

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

= 4 \cdot \frac{1}{4} (- cos (u))

Substitute back

u = e^{4 x}

= 4 \cdot \frac{1}{4} (- cos (e^{4 x}))

Simplify

4 \cdot \frac{1}{4} (- cos (e^{4 x})) : - cos (e^{4 x})

= - cos (e^{4 x})

Add a constant to the solution

= - cos (e^{4 x}) + C

\frac{d}{d x} (\frac{x}{3 x + 1})

\frac{d}{d x} (7 x + x^{2})

\frac{d}{d x} ((2 x^{4} - x + 1) (- x^{5} + 7))

\int_{0}^{x} t^{2} d t

h (x) = (- 2 x - 3)^{5} (3 x + 1)^{3}

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