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

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

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

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

0

Solution steps

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

Compute the indefinite integral:

\int cos (t) sin^{2} (t) d t = \frac{sin ^{3} ( t )}{3} + C

Compute the boundaries:

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

= 0 - 0

Simplify

= 0

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