What is the integral of xcos(8x^2+pi) ?

The integral of xcos(8x^2+pi) is -1/16 sin(8x^2)+C

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integral of xcos(8x^2+pi)

\int x cos (8 x^{2} + π) d x

- \frac{1}{16} sin (8 x^{2}) + C

Solution steps

\int x cos (8 x^{2} + π) d x

Rewrite using trig identities

= \int - x cos (8 x^{2}) d x

Take the constant out:

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

= - \int x cos (8 x^{2}) d x

Apply u-substitution

: \int \frac{cos ( u )}{16} d u

= - \int \frac{cos ( u )}{16} d u

Take the constant out:

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

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

Use the common integral:

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

= - \frac{1}{16} sin (u)

Substitute back

u = 8 x^{2}

= - \frac{1}{16} sin (8 x^{2})

Add a constant to the solution

= - \frac{1}{16} sin (8 x^{2}) + C

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What is the integral of xcos(8x^2+pi) ?
The integral of xcos(8x^2+pi) is -1/16 sin(8x^2)+C