What is the integral of sin(r^2)r ?

The integral of sin(r^2)r is -1/2 cos(r^2)+C

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integral of sin(r^2)r

\int sin (r^{2}) r d r

- \frac{1}{2} cos (r^{2}) + C

Solution steps

\int sin (r^{2}) r d r

Apply u-substitution

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

= \int \frac{sin ( 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 sin (u) d u

Use the common integral:

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

= \frac{1}{2} (- cos (u))

Substitute back

u = r^{2}

= \frac{1}{2} (- cos (r^{2}))

Simplify

= - \frac{1}{2} cos (r^{2})

Add a constant to the solution

= - \frac{1}{2} cos (r^{2}) + C

\int \frac{6 ^{x} - 4 \cdot 1 5 ^{x}}{3 ^{2 x}} d x

ln (ln (6 s))

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

\int \frac{k x}{2} - x^{2} d x

x \to \infty lim (0.5 x^{2})

What is the integral of sin(r^2)r ?
The integral of sin(r^2)r is -1/2 cos(r^2)+C