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

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

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

\int_{0}^{π} cos (x + y + z) d x

sin (y + z + π) - sin (y + z)

Solution steps

\int_{0}^{π} cos (x + y + z) d x

Apply u-substitution

: \int_{y + z}^{y + z + π} cos (u) d u

= \int_{y + z}^{y + z + π} cos (u) d u

Use the common integral:

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

= [sin (u)]_{y + z}^{y + z + π}

Compute the boundaries:

sin (y + z + π) - sin (y + z)

= sin (y + z + π) - sin (y + z)

\int_{0}^{2} (\frac{x}{2}) d x

\int_{0}^{1} (5 x^{e} + 7 e^{x}) d x

\int_{0}^{2} (3 x^{2} - x + 1) d x

\int_{0}^{4} (x - 3 x) d x

\int_{0}^{1} 2 d x

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