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

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

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

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

0

Solution steps

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

Rewrite using trig identities

= \int_{0}^{π} - 1 + 2 cos^{2} (t) + sin (2 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}^{π} 1 d t + \int_{0}^{π} 2 cos^{2} (t) d t + \int_{0}^{π} sin (2 t) d t

\int_{0}^{π} 1 d t = π

\int_{0}^{π} 2 cos^{2} (t) d t = π

\int_{0}^{π} sin (2 t) d t = 0

= - π + π + 0

Simplify

= 0

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