What is the integral from 0 to pi of |cos(x+y)| ?

The integral from 0 to pi of |cos(x+y)| is -sin(y+pi)+sin(y)

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integral from 0 to pi of |cos(x+y)|

\int_{0}^{π} ∣ cos (x + y) ∣ d x

- sin (y + π) + sin (y)

Solution steps

\int_{0}^{π} ∣ cos (x + y) ∣ d x

Eliminate Absolutes

= \int_{0}^{π} - cos (y + x) d x

Take the constant out:

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

= - \int_{0}^{π} cos (y + x) d x

Apply u-substitution

: \int_{y}^{y + π} cos (u) d u

= - \int_{y}^{y + π} cos (u) d u

Use the common integral:

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

= - [sin (u)]_{y}^{y + π}

Compute the boundaries:

sin (y + π) - sin (y)

= - (sin (y + π) - sin (y))

Simplify

= - sin (y + π) + sin (y)

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