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

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

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

\int_{0}^{2 π} sin (x) - cos (x) d x

0

Solution steps

\int_{0}^{2 π} sin (x) - cos (x) d x

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 (x) d x - \int_{0}^{2 π} cos (x) d x

\int_{0}^{2 π} sin (x) d x = 0

\int_{0}^{2 π} cos (x) d x = 0

= 0 - 0

Simplify

= 0

\int_{1}^{16} x x d x

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\int_{- π}^{π} sin^{2} (2 x) d x

\int_{1}^{2} \frac{1}{x ^{2} + 1} d x

\int_{0}^{4} 2 x + 1 d x

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