What is the integral of xcos(3pix^2) ?

The integral of xcos(3pix^2) is 1/(6pi)sin(3pix^2)+C

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integral of xcos(3pix^2)

\int x cos (3 π x^{2}) d x

\frac{1}{6 π} sin (3 π x^{2}) + C

Solution steps

\int x cos (3 π x^{2}) d x

Apply u-substitution

: \int \frac{cos ( 3 π u )}{2} d u

= \int \frac{cos ( 3 π 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 cos (3 π u) d u

Apply u-substitution

: \int cos (v) \frac{1}{3 π} d v

= \frac{1}{2} \cdot \int cos (v) \frac{1}{3 π} d v

Take the constant out:

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

= \frac{1}{2} \cdot \frac{1}{3 π} \cdot \int cos (v) d v

Use the common integral:

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

= \frac{1}{2} \cdot \frac{1}{3 π} sin (v)

Substitute back

= \frac{1}{2} \cdot \frac{1}{3 π} sin (3 π x^{2})

Simplify

\frac{1}{2} \cdot \frac{1}{3 π} sin (3 π x^{2}) : \frac{1}{6 π} sin (3 π x^{2})

= \frac{1}{6 π} sin (3 π x^{2})

Add a constant to the solution

= \frac{1}{6 π} sin (3 π x^{2}) + C

ln (x) y^{^{'}} = \frac{y}{x}

f (x) = - x, at x = - 4

y^{^{''}} - 8 y^{^{'}} = 4 t

\int \frac{2 x}{x ^{2} - 16} d x

\frac{d y}{d x}, x^{2} + y^{2} = 0

What is the integral of xcos(3pix^2) ?
The integral of xcos(3pix^2) is 1/(6pi)sin(3pix^2)+C