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

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

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

\int_{0}^{π} 2 cos (x) sin (x) d x

0

Solution steps

\int_{0}^{π} 2 cos (x) sin (x) d x

Take the constant out:

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

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

Rewrite using trig identities

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

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}^{π} sin (2 x) d x

Apply u-substitution

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

= 2 \cdot \frac{1}{2} \cdot \int_{0}^{2 π} sin (u) \frac{1}{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 \frac{1}{2} \cdot \int_{0}^{2 π} sin (u) d u

Use the common integral:

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

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

Simplify

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

= \frac{1}{2} [- cos (u)]_{0}^{2 π}

Compute the boundaries:

0

= \frac{1}{2} \cdot 0

Simplify

= 0

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