What is the integral of xcos(7pix^2) ?

The integral of xcos(7pix^2) is 1/(14pi)sin(7pix^2)+C

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integral of xcos(7pix^2)

\int x cos (7 π x^{2}) d x

\frac{1}{14 π} sin (7 π x^{2}) + C

Solution steps

\int x cos (7 π x^{2}) d x

Apply u-substitution

: \int \frac{cos ( 7 π u )}{2} d u

= \int \frac{cos ( 7 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 (7 π u) d u

Apply u-substitution

: \int cos (v) \frac{1}{7 π} d v

= \frac{1}{2} \cdot \int cos (v) \frac{1}{7 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}{7 π} \cdot \int cos (v) d v

Use the common integral:

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

= \frac{1}{2} \cdot \frac{1}{7 π} sin (v)

Substitute back

= \frac{1}{2} \cdot \frac{1}{7 π} sin (7 π x^{2})

Simplify

\frac{1}{2} \cdot \frac{1}{7 π} sin (7 π x^{2}) : \frac{1}{14 π} sin (7 π x^{2})

= \frac{1}{14 π} sin (7 π x^{2})

Add a constant to the solution

= \frac{1}{14 π} sin (7 π x^{2}) + C

- (y + t) = \frac{d y}{d t}

\int 3^{5 x} d x

3 y^{^{''}} + 10 y^{^{'}} = 0

\frac{d y}{d x} = x^{6} y^{- 3}

t^{2} \frac{d y}{d t} - t = 1 + y + t y, y (1) = 5

What is the integral of xcos(7pix^2) ?
The integral of xcos(7pix^2) is 1/(14pi)sin(7pix^2)+C