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

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

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

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

0

Solution steps

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

Compute the indefinite integral:

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

Compute the boundaries:

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

= 0 - 0

Simplify

= 0

\int_{0}^{2} (x + 2)^{3} d x

\int_{0}^{4} (x - \frac{x}{2}) d x

\int_{- 2}^{2} (x^{2} + 4) d x

\int_{- 2}^{2} x 25 - x^{2} d x

\int_{0}^{4} (x + 2) d x

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