What is the integral from 0 to 1 of-2tcos(-t^2) ?

The integral from 0 to 1 of-2tcos(-t^2) is -sin(1)

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integral from 0 to 1 of-2tcos(-t^2)

\int_{0}^{1} - 2 t cos (- t^{2}) d t

- sin (1)

Decimal

- 0.84147 \dots

Solution steps

\int_{0}^{1} - 2 t cos (- t^{2}) d t

Take the constant out:

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

= - 2 \cdot \int_{0}^{1} t cos (- t^{2}) d t

Apply u-substitution

: \int_{0}^{1} \frac{cos ( u )}{2} d u

= - 2 \cdot \int_{0}^{1} \frac{cos ( u )}{2} d u

Take the constant out:

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

= - 2 \cdot \frac{1}{2} \cdot \int_{0}^{1} cos (u) d u

Use the common integral:

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

= - 2 \cdot \frac{1}{2} [sin (u)]_{0}^{1}

Simplify

- 2 \cdot \frac{1}{2} [sin (u)]_{0}^{1} : - [sin (u)]_{0}^{1}

= - [sin (u)]_{0}^{1}

Compute the boundaries:

sin (1)

= - sin (1)

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What is the integral from 0 to 1 of-2tcos(-t^2) ?
The integral from 0 to 1 of-2tcos(-t^2) is -sin(1)