What is the integral from 0 to pi of 18cos(6x)+6yz^2 ?

The integral from 0 to pi of 18cos(6x)+6yz^2 is 6piyz^2

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integral from 0 to pi of 18cos(6x)+6yz^2

\int_{0}^{π} 18 cos (6 x) + 6 y z^{2} d x

6 π y z^{2}

Solution steps

\int_{0}^{π} 18 cos (6 x) + 6 y z^{2} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{0}^{π} 18 cos (6 x) d x + \int_{0}^{π} 6 y z^{2} d x

\int_{0}^{π} 18 cos (6 x) d x = 0

\int_{0}^{π} 6 y z^{2} d x = 6 π y z^{2}

= 0 + 6 π y z^{2}

Simplify

= 6 π y z^{2}

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What is the integral from 0 to pi of 18cos(6x)+6yz^2 ?
The integral from 0 to pi of 18cos(6x)+6yz^2 is 6piyz^2