What is the integral of xcos(6x^2) ?

The integral of xcos(6x^2) is 1/12 sin(6x^2)+C

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integral of xcos(6x^2)

\int x cos (6 x^{2}) d x

\frac{1}{12} sin (6 x^{2}) + C

Solution steps

\int x cos (6 x^{2}) d x

Apply u-substitution

: \int \frac{cos ( u )}{12} d u

= \int \frac{cos ( u )}{12} d u

Take the constant out:

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

= \frac{1}{12} \cdot \int cos (u) d u

Use the common integral:

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

= \frac{1}{12} sin (u)

Substitute back

u = 6 x^{2}

= \frac{1}{12} sin (6 x^{2})

Add a constant to the solution

= \frac{1}{12} sin (6 x^{2}) + C

\int x^{3.4} d x

y^{^{'}} = (\frac{x ^{2}}{y}), y (1) = 2

\int \frac{sin ( 4 x )}{4} d x

\frac{d}{d x} ((x^{2} - 2 x - 1) (\frac{x + 1}{x + 3}))

\frac{d y}{d t} = - \frac{2 t}{y}

What is the integral of xcos(6x^2) ?
The integral of xcos(6x^2) is 1/12 sin(6x^2)+C