What is the integral from 0 to pi/2 of-4cos(t)sin(t) ?

The integral from 0 to pi/2 of-4cos(t)sin(t) is -2

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integral from 0 to pi/2 of-4cos(t)sin(t)

\int_{0}^{\frac{π}{2}} - 4 cos (t) sin (t) d t

- 2

Solution steps

\int_{0}^{\frac{π}{2}} - 4 cos (t) sin (t) d t

Take the constant out:

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

= - 4 \cdot \int_{0}^{\frac{π}{2}} cos (t) sin (t) d t

Apply u-substitution

: \int_{0}^{1} u d u

= - 4 \cdot \int_{0}^{1} u d u

Apply the Power Rule

: [\frac{u ^{2}}{2}]_{0}^{1}

= - 4 [\frac{u ^{2}}{2}]_{0}^{1}

Compute the boundaries:

\frac{1}{2}

= - 4 \cdot \frac{1}{2}

Simplify

= - 2

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What is the integral from 0 to pi/2 of-4cos(t)sin(t) ?
The integral from 0 to pi/2 of-4cos(t)sin(t) is -2