What is the integral from 0 to pi of-cos(2y)+cos(y) ?

The integral from 0 to pi of-cos(2y)+cos(y) is 0

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integral from 0 to pi of-cos(2y)+cos(y)

\int_{0}^{π} - cos (2 y) + cos (y) d y

0

Solution steps

\int_{0}^{π} - cos (2 y) + cos (y) d y

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} cos (2 y) d y + \int_{0}^{π} cos (y) d y

\int_{0}^{π} cos (2 y) d y = 0

\int_{0}^{π} cos (y) d y = 0

= - 0 + 0

Simplify

= 0

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\int_{- 1}^{1} - 60 x^{2} d x

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What is the integral from 0 to pi of-cos(2y)+cos(y) ?
The integral from 0 to pi of-cos(2y)+cos(y) is 0