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

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

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integral of 1/2 xcos(x^2)

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

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

Solution steps

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

Apply u-substitution

: \int \frac{cos ( u )}{2} d u

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

Use the common integral:

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

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

Substitute back

u = x^{2}

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

Simplify

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

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

Add a constant to the solution

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

\int \frac{3}{x ^{3}} d x

f (t) = cos (5 t) + sin (2 t)

y^{^{''}} - y = 0, y (0) = 6

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

x \to - 1 lim (\frac{x ^{3} + 1}{x - 1})

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