What is the integral from pi to 6pi of 9cos(2x) ?

The integral from pi to 6pi of 9cos(2x) is 0

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integral from pi to 6pi of 9cos(2x)

\int_{π}^{6 π} 9 cos (2 x) d x

0

Solution steps

\int_{π}^{6 π} 9 cos (2 x) d x

Take the constant out:

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

= 9 \cdot \int_{π}^{6 π} cos (2 x) d x

Apply u-substitution

: \int_{2 π}^{12 π} cos (u) \frac{1}{2} d u

= 9 \cdot \int_{2 π}^{12 π} cos (u) \frac{1}{2} d u

Take the constant out:

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

= 9 \cdot \frac{1}{2} \cdot \int_{2 π}^{12 π} cos (u) d u

Use the common integral:

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

= 9 \cdot \frac{1}{2} [sin (u)]_{2 π}^{12 π}

Simplify

9 \cdot \frac{1}{2} [sin (u)]_{2 π}^{12 π} : \frac{9}{2} [sin (u)]_{2 π}^{12 π}

= \frac{9}{2} [sin (u)]_{2 π}^{12 π}

Compute the boundaries:

0

= \frac{9}{2} \cdot 0

Simplify

= 0

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What is the integral from pi to 6pi of 9cos(2x) ?
The integral from pi to 6pi of 9cos(2x) is 0