What is the integral of 1/2 xcos(16pix^2) ?

The integral of 1/2 xcos(16pix^2) is 1/(64pi)sin(16pix^2)+C

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integral of 1/2 xcos(16pix^2)

\int \frac{1}{2} x cos (16 π x^{2}) d x

\frac{1}{64 π} sin (16 π x^{2}) + C

Solution steps

\int \frac{1}{2} x cos (16 π x^{2}) d x

Take the constant out:

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

= \frac{1}{2} \cdot \int x cos (16 π x^{2}) d x

Apply u-substitution

: \int \frac{cos ( 16 π u )}{2} d u

= \frac{1}{2} \cdot \int \frac{cos ( 16 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 \frac{1}{2} \cdot \int cos (16 π u) d u

Apply u-substitution

: \int cos (v) \frac{1}{16 π} d v

= \frac{1}{2} \cdot \frac{1}{2} \cdot \int cos (v) \frac{1}{16 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}{2} \cdot \frac{1}{16 π} \cdot \int cos (v) d v

Use the common integral:

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

= \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{16 π} sin (v)

Substitute back

= \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{16 π} sin (16 π x^{2})

Simplify

\frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{16 π} sin (16 π x^{2}) : \frac{1}{64 π} sin (16 π x^{2})

= \frac{1}{64 π} sin (16 π x^{2})

Add a constant to the solution

= \frac{1}{64 π} sin (16 π x^{2}) + C

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What is the integral of 1/2 xcos(16pix^2) ?
The integral of 1/2 xcos(16pix^2) is 1/(64pi)sin(16pix^2)+C