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

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

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

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

0

Solution steps

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

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

\int_{0}^{2 π} sin (t) d t = 0

\int_{0}^{2 π} cos (t) d t = 0

= - 0 + 0

Simplify

= 0

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