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

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

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

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

0

Solution steps

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

Compute the indefinite integral:

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

Compute the boundaries:

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

= 0 - 0

Simplify

= 0

\int_{0}^{x} 1 - x d x

\int_{2}^{4} \frac{m ^{2}}{m + 1} d m

\int_{- 1}^{5} 4 x - x^{2} + 5 d x

\int_{0}^{π} (x^{2}) cos (n x) d x

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

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