What is the integral of 5e^{6x}cos(e^{6x}) ?

The integral of 5e^{6x}cos(e^{6x}) is 5/6 sin(e^{6x})+C

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integral of 5e^{6x}cos(e^{6x})

\int 5 e^{6 x} cos (e^{6 x}) d x

\frac{5}{6} sin (e^{6 x}) + C

Solution steps

\int 5 e^{6 x} cos (e^{6 x}) d x

Take the constant out:

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

= 5 \cdot \int e^{6 x} cos (e^{6 x}) d x

Apply u-substitution

: \int e^{u} cos (e^{u}) \frac{1}{6} d u

= 5 \cdot \int e^{u} cos (e^{u}) \frac{1}{6} d u

Take the constant out:

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

= 5 \cdot \frac{1}{6} \cdot \int e^{u} cos (e^{u}) d u

Apply u-substitution

: \int cos (v) d v

= 5 \cdot \frac{1}{6} \cdot \int cos (v) d v

Use the common integral:

\int cos (v) d v = sin (v)

= 5 \cdot \frac{1}{6} sin (v)

Substitute back

= 5 \cdot \frac{1}{6} sin (e^{6 x})

Simplify

5 \cdot \frac{1}{6} sin (e^{6 x}) : \frac{5}{6} sin (e^{6 x})

= \frac{5}{6} sin (e^{6 x})

Add a constant to the solution

= \frac{5}{6} sin (e^{6 x}) + C

y^{^{''}} + 2 y^{^{'}} + 5 y = 4 cos (2 t) e^{- t}

\frac{\partial}{\partial x} (sin (x^{2} - 9 x y))

\int \frac{1}{cos ^{5} ( x )} d x

2 y y^{^{'}} + 2 = y^{2} + 2 x, y (0) = 6

\frac{2}{x ^{\frac{1}{3}}} - 2

What is the integral of 5e^{6x}cos(e^{6x}) ?
The integral of 5e^{6x}cos(e^{6x}) is 5/6 sin(e^{6x})+C