What is the integral from 0 to pi of cos(u+v+w) ?

The integral from 0 to pi of cos(u+v+w) is sin(v+w+pi)-sin(v+w)

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integral from 0 to pi of cos(u+v+w)

\int_{0}^{π} cos (u + v + w) d u

sin (v + w + π) - sin (v + w)

Solution steps

\int_{0}^{π} cos (u + v + w) d u

Apply u-substitution

: \int_{v + w}^{v + w + π} cos (t) d t

= \int_{v + w}^{v + w + π} cos (t) d t

Use the common integral:

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

= [sin (t)]_{v + w}^{v + w + π}

Compute the boundaries:

sin (v + w + π) - sin (v + w)

= sin (v + w + π) - sin (v + w)

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\int_{0}^{8} (x - \frac{1}{2}) d x

\int_{0}^{1} x^{2} - 3 x + 5 d x

\int_{0}^{1} x^{2} - 3 x + 6 d x

\int_{0}^{2 x} e^{- x^{2}} d x

What is the integral from 0 to pi of cos(u+v+w) ?
The integral from 0 to pi of cos(u+v+w) is sin(v+w+pi)-sin(v+w)