What is the integral of t*sin(t^2) ?

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

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integral of t*sin(t^2)

\int t \cdot sin (t^{2}) d t

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

Solution steps

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

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 = t^{2}

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

Simplify

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

Add a constant to the solution

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

\frac{d}{d x} (- sin (x - 2))

n = 0 \sum \infty - 6 \cdot (- \frac{1}{3})^{n}

\frac{d}{d x} (x^{3} - 11 x^{2} + 31 x - 21)

\frac{d y}{d x} = \frac{y ^{2} - x ^{2}}{x y}

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