What is the integral of 1/2 xcos(4x^2) ?

The integral of 1/2 xcos(4x^2) is 1/16 sin(4x^2)+C

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integral of 1/2 xcos(4x^2)

\int \frac{1}{2} x cos (4 x^{2}) d x

\frac{1}{16} sin (4 x^{2}) + C

Solution steps

\int \frac{1}{2} x cos (4 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 (4 x^{2}) d x

Apply u-substitution

: \int \frac{cos ( u )}{8} d u

= \frac{1}{2} \cdot \int \frac{cos ( u )}{8} 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}{8} \cdot \int cos (u) d u

Use the common integral:

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

= \frac{1}{2} \cdot \frac{1}{8} sin (u)

Substitute back

u = 4 x^{2}

= \frac{1}{2} \cdot \frac{1}{8} sin (4 x^{2})

Simplify

\frac{1}{2} \cdot \frac{1}{8} sin (4 x^{2}) : \frac{1}{16} sin (4 x^{2})

= \frac{1}{16} sin (4 x^{2})

Add a constant to the solution

= \frac{1}{16} sin (4 x^{2}) + C

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\frac{\partial}{\partial y} (ln (1 + x y))

n = 1 \sum \infty \frac{1}{n} - \frac{1}{2 n + 1}

1 - x^{2}, 0

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

What is the integral of 1/2 xcos(4x^2) ?
The integral of 1/2 xcos(4x^2) is 1/16 sin(4x^2)+C