What is the integral from 0 to pi/2 of 8sin(x)cos(x) ?

The integral from 0 to pi/2 of 8sin(x)cos(x) is 4

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integral from 0 to pi/2 of 8sin(x)cos(x)

\int_{0}^{\frac{π}{2}} 8 sin (x) cos (x) d x

4

Solution steps

\int_{0}^{\frac{π}{2}} 8 sin (x) cos (x) d x

Take the constant out:

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

= 8 \cdot \int_{0}^{\frac{π}{2}} sin (x) cos (x) d x

Apply u-substitution

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

= 8 \cdot \int_{0}^{1} u d u

Apply the Power Rule

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

= 8 [\frac{u ^{2}}{2}]_{0}^{1}

Compute the boundaries:

\frac{1}{2}

= 8 \cdot \frac{1}{2}

Simplify

= 4

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What is the integral from 0 to pi/2 of 8sin(x)cos(x) ?
The integral from 0 to pi/2 of 8sin(x)cos(x) is 4